[proofplan] We verify the Lie algebra axioms directly from the defining bracket. Bilinearity follows by expanding the formula, using bilinearity in $\mathfrak h$ and $\mathfrak k$, and using the $F$-linearity of the action map $\rho:\mathfrak k\to\operatorname{Der}_F(\mathfrak h)$. Alternation is checked directly by evaluating $[(h,k),(h,k)]_{\rtimes}$, which also gives skew-symmetry by bilinearity. The Jacobi identity is checked componentwise: the $\mathfrak k$-component is the Jacobi identity in $\mathfrak k$, while the $\mathfrak h$-component splits into the Jacobi identity in $\mathfrak h$, the derivation rule for each $\rho(k)$, and the homomorphism identity $\rho([k_i,k_j]_{\mathfrak k})=[\rho(k_i),\rho(k_j)]$. Finally, the two canonical embeddings are inspected directly from the bracket formula. [/proofplan]

[step:Verify bilinearity and alternation of the bracket] Let $x_1=(h_1,k_1)$ and $x_2=(h_2,k_2)$ be elements of $\mathfrak h \oplus \mathfrak k$. Since $[\cdot,\cdot]_{\mathfrak h}$ and $[\cdot,\cdot]_{\mathfrak k}$ are $F$-bilinear, each $\rho(k):\mathfrak h \to \mathfrak h$ is $F$-linear, and $\rho:\mathfrak k\to\operatorname{Der}_F(\mathfrak h)$ is $F$-linear, the formula defining $[\cdot,\cdot]_{\rtimes}$ is $F$-bilinear in $x_1$ and $x_2$. For alternation, let $x=(h,k) \in \mathfrak h \oplus \mathfrak k$. Since $\mathfrak h$ and $\mathfrak k$ are Lie algebras, their brackets are alternating, so $[h,h]_{\mathfrak h}=0$ and $[k,k]_{\mathfrak k}=0$. Therefore \begin{align*} [x,x]_{\rtimes} &= [(h,k),(h,k)]_{\rtimes} \\ &= \bigl([h,h]_{\mathfrak h}+\rho(k)(h)-\rho(k)(h),[k,k]_{\mathfrak k}\bigr) \\ &= (0,0). \end{align*} Thus $[\cdot,\cdot]_{\rtimes}$ is alternating. Since the bracket is bilinear, alternation also implies skew-symmetry: applying alternation to $x_1+x_2$ gives \begin{align*} 0 &=[x_1+x_2,x_1+x_2]_{\rtimes} \\ &=[x_1,x_2]_{\rtimes}+[x_2,x_1]_{\rtimes}, \end{align*} and hence $[x_2,x_1]_{\rtimes}=-[x_1,x_2]_{\rtimes}$. [/step]

[step:Expand the Jacobi expression componentwise]Let $x_i=(h_i,k_i) \in \mathfrak h \oplus \mathfrak k$ for $i \in \{1,2,3\}$. Define $D_i := \rho(k_i) \in \operatorname{Der}_F(\mathfrak h)$ for each $i$. Since $\rho$ is a Lie algebra homomorphism, \begin{align*} [D_i,D_j] = D_iD_j-D_jD_i = \rho([k_i,k_j]_{\mathfrak k}) \end{align*} for all $i,j \in \{1,2,3\}$. Since $D_i$ is a derivation of $\mathfrak h$, \begin{align*} D_i([a,b]_{\mathfrak h})=[D_i(a),b]_{\mathfrak h}+[a,D_i(b)]_{\mathfrak h} \end{align*} for all $a,b \in \mathfrak h$. Set \begin{align*} J(x_1,x_2,x_3) := [x_1,[x_2,x_3]_{\rtimes}]_{\rtimes} +[x_2,[x_3,x_1]_{\rtimes}]_{\rtimes} +[x_3,[x_1,x_2]_{\rtimes}]_{\rtimes}. \end{align*} The $\mathfrak k$-component of $J(x_1,x_2,x_3)$ is \begin{align*} [k_1,[k_2,k_3]_{\mathfrak k}]_{\mathfrak k} +[k_2,[k_3,k_1]_{\mathfrak k}]_{\mathfrak k} +[k_3,[k_1,k_2]_{\mathfrak k}]_{\mathfrak k}, \end{align*} which is $0$ by the Jacobi identity in $\mathfrak k$.[/step]

[guided]We check the Jacobi identity for three arbitrary elements $x_i=(h_i,k_i) \in \mathfrak h \oplus \mathfrak k$, where $i \in \{1,2,3\}$. The bracket has two components, so the Jacobi expression can vanish only if both its $\mathfrak h$-component and its $\mathfrak k$-component vanish. Define $D_i := \rho(k_i) \in \operatorname{Der}_F(\mathfrak h)$ for each $i$. This notation records the action of the $\mathfrak k$-part of $x_i$ on $\mathfrak h$. Because $\rho$ is a Lie algebra homomorphism, the commutator of the two derivations $D_i$ and $D_j$ is the derivation attached to the bracket $[k_i,k_j]_{\mathfrak k}$: \begin{align*} [D_i,D_j]=D_iD_j-D_jD_i=\rho([k_i,k_j]_{\mathfrak k}). \end{align*} Because $D_i$ is a derivation of the Lie algebra $\mathfrak h$, it satisfies \begin{align*} D_i([a,b]_{\mathfrak h})=[D_i(a),b]_{\mathfrak h}+[a,D_i(b)]_{\mathfrak h} \end{align*} for all $a,b \in \mathfrak h$. Now define the Jacobi expression \begin{align*} J(x_1,x_2,x_3) := [x_1,[x_2,x_3]_{\rtimes}]_{\rtimes} +[x_2,[x_3,x_1]_{\rtimes}]_{\rtimes} +[x_3,[x_1,x_2]_{\rtimes}]_{\rtimes}. \end{align*} The second component of $[x_i,x_j]_{\rtimes}$ is exactly $[k_i,k_j]_{\mathfrak k}$. Therefore the $\mathfrak k$-component of $J(x_1,x_2,x_3)$ is \begin{align*} [k_1,[k_2,k_3]_{\mathfrak k}]_{\mathfrak k} +[k_2,[k_3,k_1]_{\mathfrak k}]_{\mathfrak k} +[k_3,[k_1,k_2]_{\mathfrak k}]_{\mathfrak k}. \end{align*} This is $0$ by the Jacobi identity in the Lie algebra $\mathfrak k$. Thus only the $\mathfrak h$-component remains to be checked.[/guided]

[step:Cancel the $\mathfrak h$ component of the Jacobi expression]The $\mathfrak h$-component of $[x_1,[x_2,x_3]_{\rtimes}]_{\rtimes}$ equals \begin{align*} &[h_1,[h_2,h_3]_{\mathfrak h}+D_2(h_3)-D_3(h_2)]_{\mathfrak h} +D_1([h_2,h_3]_{\mathfrak h}+D_2(h_3)-D_3(h_2)) \\ &\qquad -\rho([k_2,k_3]_{\mathfrak k})(h_1). \end{align*} Using $\rho([k_2,k_3]_{\mathfrak k})=[D_2,D_3]$ and expanding by bilinearity and the derivation rule gives \begin{align*} &[h_1,[h_2,h_3]_{\mathfrak h}]_{\mathfrak h} +[h_1,D_2(h_3)]_{\mathfrak h} -[h_1,D_3(h_2)]_{\mathfrak h} \\ &\quad +[D_1(h_2),h_3]_{\mathfrak h} +[h_2,D_1(h_3)]_{\mathfrak h} +D_1D_2(h_3)-D_1D_3(h_2)-D_2D_3(h_1)+D_3D_2(h_1). \end{align*} Adding the two cyclic analogues, the terms \begin{align*} [h_1,[h_2,h_3]_{\mathfrak h}]_{\mathfrak h} +[h_2,[h_3,h_1]_{\mathfrak h}]_{\mathfrak h} +[h_3,[h_1,h_2]_{\mathfrak h}]_{\mathfrak h} \end{align*} cancel by the Jacobi identity in $\mathfrak h$. The terms involving one bracket in $\mathfrak h$ and one derivation are \begin{align*} &[h_1,D_2(h_3)]_{\mathfrak h}-[h_1,D_3(h_2)]_{\mathfrak h}+[D_1(h_2),h_3]_{\mathfrak h}+[h_2,D_1(h_3)]_{\mathfrak h} \\ &\quad +[h_2,D_3(h_1)]_{\mathfrak h}-[h_2,D_1(h_3)]_{\mathfrak h}+[D_2(h_3),h_1]_{\mathfrak h}+[h_3,D_2(h_1)]_{\mathfrak h} \\ &\quad +[h_3,D_1(h_2)]_{\mathfrak h}-[h_3,D_2(h_1)]_{\mathfrak h}+[D_3(h_1),h_2]_{\mathfrak h}+[h_1,D_3(h_2)]_{\mathfrak h}. \end{align*} They cancel pairwise: the pairs $[h_1,D_2(h_3)]_{\mathfrak h}+[D_2(h_3),h_1]_{\mathfrak h}$, $[D_1(h_2),h_3]_{\mathfrak h}+[h_3,D_1(h_2)]_{\mathfrak h}$, and $[h_2,D_3(h_1)]_{\mathfrak h}+[D_3(h_1),h_2]_{\mathfrak h}$ vanish by skew-symmetry in $\mathfrak h$, while the remaining three pairs cancel as identical terms with opposite signs. The second-order derivation terms are \begin{align*} &D_1D_2(h_3)-D_1D_3(h_2)-D_2D_3(h_1)+D_3D_2(h_1) \\ &\quad +D_2D_3(h_1)-D_2D_1(h_3)-D_3D_1(h_2)+D_1D_3(h_2) \\ &\quad +D_3D_1(h_2)-D_3D_2(h_1)-D_1D_2(h_3)+D_2D_1(h_3), \end{align*} which cancel term by term. Hence the $\mathfrak h$-component of $J(x_1,x_2,x_3)$ is $0$.[/step]

[guided]We now prove that the $\mathfrak h$-component of the Jacobi expression also vanishes. First compute the first cyclic term. Since \begin{align*} [x_2,x_3]_{\rtimes} = \bigl([h_2,h_3]_{\mathfrak h}+D_2(h_3)-D_3(h_2),[k_2,k_3]_{\mathfrak k}\bigr), \end{align*} the definition of the semidirect bracket gives the $\mathfrak h$-component of $[x_1,[x_2,x_3]_{\rtimes}]_{\rtimes}$ as \begin{align*} &[h_1,[h_2,h_3]_{\mathfrak h}+D_2(h_3)-D_3(h_2)]_{\mathfrak h} +D_1([h_2,h_3]_{\mathfrak h}+D_2(h_3)-D_3(h_2)) \\ &\qquad -\rho([k_2,k_3]_{\mathfrak k})(h_1). \end{align*} The last term is where the homomorphism property of $\rho$ is used. Since $\rho([k_2,k_3]_{\mathfrak k})=[D_2,D_3]=D_2D_3-D_3D_2$, this expression becomes \begin{align*} &[h_1,[h_2,h_3]_{\mathfrak h}]_{\mathfrak h} +[h_1,D_2(h_3)]_{\mathfrak h} -[h_1,D_3(h_2)]_{\mathfrak h} \\ &\quad +D_1([h_2,h_3]_{\mathfrak h}) +D_1D_2(h_3)-D_1D_3(h_2) -D_2D_3(h_1)+D_3D_2(h_1). \end{align*} Now use that $D_1$ is a derivation of $\mathfrak h$: \begin{align*} D_1([h_2,h_3]_{\mathfrak h}) = [D_1(h_2),h_3]_{\mathfrak h} +[h_2,D_1(h_3)]_{\mathfrak h}. \end{align*} Thus the first cyclic term contributes \begin{align*} &[h_1,[h_2,h_3]_{\mathfrak h}]_{\mathfrak h} +[h_1,D_2(h_3)]_{\mathfrak h} -[h_1,D_3(h_2)]_{\mathfrak h} \\ &\quad +[D_1(h_2),h_3]_{\mathfrak h} +[h_2,D_1(h_3)]_{\mathfrak h} +D_1D_2(h_3)-D_1D_3(h_2)-D_2D_3(h_1)+D_3D_2(h_1). \end{align*} We add the two cyclic analogues obtained by replacing $(1,2,3)$ with $(2,3,1)$ and $(3,1,2)$. Three different kinds of terms appear. First, the pure $\mathfrak h$-bracket terms are \begin{align*} [h_1,[h_2,h_3]_{\mathfrak h}]_{\mathfrak h} +[h_2,[h_3,h_1]_{\mathfrak h}]_{\mathfrak h} +[h_3,[h_1,h_2]_{\mathfrak h}]_{\mathfrak h}, \end{align*} and these vanish by the Jacobi identity in $\mathfrak h$. Second, the terms containing one derivation and one $\mathfrak h$-bracket are \begin{align*} &[h_1,D_2(h_3)]_{\mathfrak h}-[h_1,D_3(h_2)]_{\mathfrak h}+[D_1(h_2),h_3]_{\mathfrak h}+[h_2,D_1(h_3)]_{\mathfrak h} \\ &\quad +[h_2,D_3(h_1)]_{\mathfrak h}-[h_2,D_1(h_3)]_{\mathfrak h}+[D_2(h_3),h_1]_{\mathfrak h}+[h_3,D_2(h_1)]_{\mathfrak h} \\ &\quad +[h_3,D_1(h_2)]_{\mathfrak h}-[h_3,D_2(h_1)]_{\mathfrak h}+[D_3(h_1),h_2]_{\mathfrak h}+[h_1,D_3(h_2)]_{\mathfrak h}. \end{align*} Now pair them explicitly. The sums \begin{align*} [h_1,D_2(h_3)]_{\mathfrak h}+[D_2(h_3),h_1]_{\mathfrak h},\qquad [D_1(h_2),h_3]_{\mathfrak h}+[h_3,D_1(h_2)]_{\mathfrak h}, \end{align*} and \begin{align*} [h_2,D_3(h_1)]_{\mathfrak h}+[D_3(h_1),h_2]_{\mathfrak h} \end{align*} vanish by skew-symmetry of $[\cdot,\cdot]_{\mathfrak h}$. The remaining three cancellations are literal opposite-sign cancellations: \begin{align*} -[h_1,D_3(h_2)]_{\mathfrak h}+[h_1,D_3(h_2)]_{\mathfrak h}=0, \end{align*} \begin{align*} [h_2,D_1(h_3)]_{\mathfrak h}-[h_2,D_1(h_3)]_{\mathfrak h}=0, \end{align*} and \begin{align*} [h_3,D_2(h_1)]_{\mathfrak h}-[h_3,D_2(h_1)]_{\mathfrak h}=0. \end{align*} Third, the second-order derivation terms are \begin{align*} &D_1D_2(h_3)-D_1D_3(h_2)-D_2D_3(h_1)+D_3D_2(h_1) \\ &\quad +D_2D_3(h_1)-D_2D_1(h_3)-D_3D_1(h_2)+D_1D_3(h_2) \\ &\quad +D_3D_1(h_2)-D_3D_2(h_1)-D_1D_2(h_3)+D_2D_1(h_3). \end{align*} Each displayed term appears once with coefficient $1$ and once with coefficient $-1$, so their sum is $0$. Therefore the $\mathfrak h$-component of $J(x_1,x_2,x_3)$ is $0$.[/guided]

[step:Conclude the Jacobi identity for the semidirect bracket] The previous two steps show that both the $\mathfrak h$-component and the $\mathfrak k$-component of $J(x_1,x_2,x_3)$ vanish. Hence \begin{align*} [x_1,[x_2,x_3]_{\rtimes}]_{\rtimes} +[x_2,[x_3,x_1]_{\rtimes}]_{\rtimes} +[x_3,[x_1,x_2]_{\rtimes}]_{\rtimes} =0 \end{align*} for all $x_1,x_2,x_3 \in \mathfrak h \oplus \mathfrak k$. Together with bilinearity and alternation, this proves that $\mathfrak h \rtimes_\rho \mathfrak k$ is a Lie algebra over $F$. [/step]

[step:Identify $\mathfrak h$ as an ideal and $\mathfrak k$ as a subalgebra] Define \begin{align*} i_{\mathfrak h}: \mathfrak h &\to \mathfrak h \rtimes_\rho \mathfrak k, & h &\mapsto (h,0), \\ i_{\mathfrak k}: \mathfrak k &\to \mathfrak h \rtimes_\rho \mathfrak k, & k &\mapsto (0,k). \end{align*} Both maps are injective $F$-linear maps. For $h,h' \in \mathfrak h$, \begin{align*} [i_{\mathfrak h}(h),i_{\mathfrak h}(h')]_{\rtimes} = [(h,0),(h',0)]_{\rtimes} = ([h,h']_{\mathfrak h},0) = i_{\mathfrak h}([h,h']_{\mathfrak h}), \end{align*} so $i_{\mathfrak h}(\mathfrak h)$ is a Lie subalgebra. For any $(a,k) \in \mathfrak h \rtimes_\rho \mathfrak k$ and any $h \in \mathfrak h$, \begin{align*} [(a,k),(h,0)]_{\rtimes} = ([a,h]_{\mathfrak h}+\rho(k)(h),0) \in i_{\mathfrak h}(\mathfrak h). \end{align*} Thus $i_{\mathfrak h}(\mathfrak h)$ is an ideal. For $k,k' \in \mathfrak k$, \begin{align*} [i_{\mathfrak k}(k),i_{\mathfrak k}(k')]_{\rtimes} = [(0,k),(0,k')]_{\rtimes} = (0,[k,k']_{\mathfrak k}) = i_{\mathfrak k}([k,k']_{\mathfrak k}). \end{align*} Therefore $i_{\mathfrak k}(\mathfrak k)$ is a Lie subalgebra of $\mathfrak h \rtimes_\rho \mathfrak k$. This proves all asserted properties of the semidirect product Lie algebra. [/step]