[proofplan]
We prove the result directly from the limit definition of [iterated directional derivative](/page/Iterated%20Directional%20Derivative). First we use linearity of ordinary limits to compute the inner [directional derivative](/page/Directional%20Derivative) in direction $v$ of the function $\alpha f+\beta g$ at every point needed for the outer derivative. Then we apply linearity of limits a second time to the one-variable difference quotient defining the outer directional derivative in direction $u$.
[/proofplan]
[step:Define the scalar-line functions for the outer derivative]
Define the function
\begin{align*}
h: U \to \mathbb{R}^n
\end{align*}
by $h(x)=\alpha f(x)+\beta g(x)$ for every $x \in U$. Since $U$ is open and $a \in U$, there exists $\rho>0$ such that $a+su \in U$ whenever $s \in \mathbb{R}$ and $|s|<\rho$.
By the meaning of the existence of $D_uD_v f(a)$ and $D_uD_v g(a)$, after possibly decreasing $\rho$, the directional derivatives $D_v f(a+su)$ and $D_v g(a+su)$ exist for every $s \in \mathbb{R}$ with $|s|<\rho$, and the limits
\begin{align*}
D_uD_v f(a)=\lim_{s\to 0}\frac{D_v f(a+su)-D_v f(a)}{s}
\end{align*}
and
\begin{align*}
D_uD_v g(a)=\lim_{s\to 0}\frac{D_v g(a+su)-D_v g(a)}{s}
\end{align*}
exist in $\mathbb{R}^n$.
Define the functions
\begin{align*}
F: (-\rho,\rho) \to \mathbb{R}^n
\end{align*}
and
\begin{align*}
G: (-\rho,\rho) \to \mathbb{R}^n
\end{align*}
by $F(s)=D_v f(a+su)$ and $G(s)=D_v g(a+su)$. Then
\begin{align*}
D_uD_v f(a)=\lim_{s\to 0}\frac{F(s)-F(0)}{s}
\end{align*}
and
\begin{align*}
D_uD_v g(a)=\lim_{s\to 0}\frac{G(s)-G(0)}{s}.
\end{align*}
[/step]
[step:Compute the inner directional derivative of the linear combination]
Fix $s \in (-\rho,\rho)$ and set $x_s=a+su$. Since $U$ is open and $x_s \in U$, the expression $x_s+tv$ lies in $U$ for all sufficiently small $t \in \mathbb{R}$. Because $D_v f(x_s)$ and $D_v g(x_s)$ exist, linearity of limits in $\mathbb{R}^n$ gives
\begin{align*}
D_v h(x_s)=\lim_{t\to 0}\frac{h(x_s+tv)-h(x_s)}{t}.
\end{align*}
Substituting $h=\alpha f+\beta g$ into the difference quotient gives
\begin{align*}
D_v h(x_s)=\lim_{t\to 0}\left(\alpha\frac{f(x_s+tv)-f(x_s)}{t}+\beta\frac{g(x_s+tv)-g(x_s)}{t}\right).
\end{align*}
Since the two limits defining $D_v f(x_s)$ and $D_v g(x_s)$ exist, linearity of limits yields
\begin{align*}
D_v h(x_s)=\alpha D_v f(x_s)+\beta D_v g(x_s).
\end{align*}
Thus, for every $s \in (-\rho,\rho)$,
\begin{align*}
D_v h(a+su)=\alpha F(s)+\beta G(s).
\end{align*}
[guided]
Fix $s \in (-\rho,\rho)$ and define the point $x_s \in U$ by $x_s=a+su$. We need to understand the inner derivative $D_v h(x_s)$ because the outer derivative $D_uD_v h(a)$ is formed by differentiating the function $x \mapsto D_v h(x)$ along the line $a+su$.
Since $U$ is open and $x_s \in U$, the points $x_s+tv$ remain in $U$ for all sufficiently small real $t$. Therefore the directional difference quotient for $h$ in direction $v$ at $x_s$ is well-defined for all sufficiently small nonzero $t$:
\begin{align*}
\frac{h(x_s+tv)-h(x_s)}{t}.
\end{align*}
Using the definition $h=\alpha f+\beta g$, this quotient is
\begin{align*}
\frac{\alpha f(x_s+tv)+\beta g(x_s+tv)-\alpha f(x_s)-\beta g(x_s)}{t}.
\end{align*}
Collecting the $f$-terms and the $g$-terms gives
\begin{align*}
\frac{h(x_s+tv)-h(x_s)}{t}=\alpha\frac{f(x_s+tv)-f(x_s)}{t}+\beta\frac{g(x_s+tv)-g(x_s)}{t}.
\end{align*}
The hypotheses giving the existence of the iterated derivatives ensure, for this $s$, that both inner directional derivatives $D_v f(x_s)$ and $D_v g(x_s)$ exist. Hence the two limits
\begin{align*}
\lim_{t\to 0}\frac{f(x_s+tv)-f(x_s)}{t}=D_v f(x_s)
\end{align*}
and
\begin{align*}
\lim_{t\to 0}\frac{g(x_s+tv)-g(x_s)}{t}=D_v g(x_s)
\end{align*}
exist in $\mathbb{R}^n$. Linearity of limits in the finite-dimensional [vector space](/page/Vector%20Space) $\mathbb{R}^n$ then gives
\begin{align*}
D_v h(x_s)=\alpha D_v f(x_s)+\beta D_v g(x_s).
\end{align*}
Since $F(s)$ and $G(s)$ were defined by $F(s)=D_v f(a+su)$ and $G(s)=D_v g(a+su)$, this is equivalently
\begin{align*}
D_v h(a+su)=\alpha F(s)+\beta G(s).
\end{align*}
[/guided]
[/step]
[step:Differentiate the inner derivative in direction $u$]
The preceding step shows that $D_v h(a+su)$ exists for every $s \in (-\rho,\rho)$ and that
\begin{align*}
D_v h(a+su)=\alpha F(s)+\beta G(s).
\end{align*}
Therefore the outer difference quotient for $h$ is
\begin{align*}
\frac{D_v h(a+su)-D_v h(a)}{s}=\alpha\frac{F(s)-F(0)}{s}+\beta\frac{G(s)-G(0)}{s}
\end{align*}
for every nonzero $s \in (-\rho,\rho)$. Taking $s \to 0$ and using the existence of the limits defining $D_uD_v f(a)$ and $D_uD_v g(a)$ gives
\begin{align*}
D_uD_v h(a)=\alpha D_uD_v f(a)+\beta D_uD_v g(a).
\end{align*}
Thus $D_uD_v h(a)$ exists.
[/step]
[step:Return to the original function]
Since $h=\alpha f+\beta g$ as a function from $U$ to $\mathbb{R}^n$, the identity just proved is exactly
\begin{align*}
D_uD_v(\alpha f+\beta g)(a)=\alpha D_uD_v f(a)+\beta D_uD_v g(a).
\end{align*}
This proves the claimed linearity of the iterated directional derivative in the function argument.
[/step]
