[proofplan]
Fix $X\in\mathfrak g$ and compare two curves in $H$: the image under $\varphi$ of the exponential one-parameter subgroup generated by $X$, and the exponential one-parameter subgroup generated by $d\varphi(X)$. The homomorphism property of $\varphi$ makes the first curve a one-parameter subgroup of $H$. The chain rule identifies its initial tangent vector with $d\varphi(X)$, so the defining uniqueness property of the Lie group exponential gives equality of the two curves for all real parameters. Setting $t=1$ then gives the stated special case.
[/proofplan]
[step:Define the image curve and verify that it is a one-parameter subgroup of $H$]
Fix $X\in\mathfrak g$. Define the smooth curve
\begin{align*}
\gamma:\mathbb R\to H,\qquad t\mapsto \varphi(\exp_G(tX)).
\end{align*}
The curve is smooth because the map $t\mapsto \exp_G(tX)$ is smooth and $\varphi$ is smooth.
For $s,t\in\mathbb R$, the exponential curve
\begin{align*}
\alpha_X:\mathbb R\to G,\qquad u\mapsto \exp_G(uX)
\end{align*}
is the one-parameter subgroup of $G$ with initial tangent vector $X$. Hence
\begin{align*}
\exp_G((s+t)X)=\exp_G(sX)\exp_G(tX).
\end{align*}
Since $\varphi$ is a [group homomorphism](/page/Group%20Homomorphism),
\begin{align*}
\gamma(s+t)=\varphi(\exp_G((s+t)X)).
\end{align*}
Using the one-parameter subgroup identity in $G$ and then the homomorphism property of $\varphi$ gives
\begin{align*}
\gamma(s+t)=\varphi(\exp_G(sX)\exp_G(tX)).
\end{align*}
Therefore
\begin{align*}
\gamma(s+t)=\varphi(\exp_G(sX))\varphi(\exp_G(tX))=\gamma(s)\gamma(t).
\end{align*}
Also,
\begin{align*}
\gamma(0)=\varphi(\exp_G(0))=\varphi(e_G)=e_H.
\end{align*}
Thus $\gamma$ is a smooth one-parameter subgroup of $H$.
[/step]
[step:Compute the initial tangent vector of the image curve]
We compute the derivative of $\gamma$ at $0$. Let
\begin{align*}
\alpha_X:\mathbb R\to G,\qquad t\mapsto \exp_G(tX).
\end{align*}
Then $\gamma=\varphi\circ\alpha_X$. Identifying $T_0\mathbb R$ with $\mathbb R$ by the standard coordinate vector $1\in T_0\mathbb R$, the defining property of the exponential gives
\begin{align*}
d(\alpha_X)_0(1)=X.
\end{align*}
By the chain rule for smooth maps,
\begin{align*}
d\gamma_0(1)=d\varphi_{e_G}(d(\alpha_X)_0(1)).
\end{align*}
Substituting $d(\alpha_X)_0(1)=X$ and $d\varphi=d\varphi_{e_G}$ yields
\begin{align*}
d\gamma_0(1)=d\varphi(X).
\end{align*}
Thus the initial tangent vector of the one-parameter subgroup $\gamma$ is $d\varphi(X)$.
[guided]
We now identify which element of $\mathfrak h=T_{e_H}H$ generates the one-parameter subgroup $\gamma$. The curve already has the form
\begin{align*}
\gamma=\varphi\circ\alpha_X,
\end{align*}
where
\begin{align*}
\alpha_X:\mathbb R\to G,\qquad t\mapsto \exp_G(tX).
\end{align*}
The exponential map on $G$ is characterized so that $\alpha_X$ is the one-parameter subgroup whose initial velocity is $X$. With the standard identification of $T_0\mathbb R$ with $\mathbb R$ through the coordinate vector $1\in T_0\mathbb R$, this means
\begin{align*}
d(\alpha_X)_0(1)=X.
\end{align*}
Because $\varphi:G\to H$ is smooth, the ordinary chain rule applies to the composition $\gamma=\varphi\circ\alpha_X$. At $0\in\mathbb R$, the point $\alpha_X(0)$ is $e_G$, so the chain rule gives
\begin{align*}
d\gamma_0(1)=d\varphi_{\alpha_X(0)}(d(\alpha_X)_0(1)).
\end{align*}
Since $\alpha_X(0)=e_G$, this becomes
\begin{align*}
d\gamma_0(1)=d\varphi_{e_G}(d(\alpha_X)_0(1)).
\end{align*}
Now substitute the defining initial-velocity identity $d(\alpha_X)_0(1)=X$:
\begin{align*}
d\gamma_0(1)=d\varphi_{e_G}(X).
\end{align*}
By the notation in the statement, $d\varphi=d\varphi_{e_G}$, so
\begin{align*}
d\gamma_0(1)=d\varphi(X).
\end{align*}
Thus the image curve $\gamma$ is not merely some one-parameter subgroup of $H$; it is precisely a one-parameter subgroup with initial tangent vector $d\varphi(X)$.
[/guided]
[/step]
[step:Use uniqueness of one-parameter subgroups with prescribed initial tangent vector]
Define the exponential one-parameter subgroup in $H$ generated by $d\varphi(X)$ by
\begin{align*}
\beta:\mathbb R\to H,\qquad t\mapsto \exp_H(t\,d\varphi(X)).
\end{align*}
By the defining property of the Lie group exponential, $\beta$ is the unique smooth one-parameter subgroup of $H$ whose initial tangent vector is $d\varphi(X)$.
The previous steps show that $\gamma$ is also a smooth one-parameter subgroup of $H$ and that its initial tangent vector is $d\varphi(X)$. Therefore uniqueness gives
\begin{align*}
\gamma(t)=\beta(t)
\end{align*}
for every $t\in\mathbb R$. Unwinding the definitions of $\gamma$ and $\beta$ gives
\begin{align*}
\varphi(\exp_G(tX))=\exp_H(t\,d\varphi(X))
\end{align*}
for every $t\in\mathbb R$.
[/step]
[step:Set $t=1$ to obtain the particular case]
Taking $t=1$ in the identity already proved gives
\begin{align*}
\varphi(\exp_G(1\cdot X))=\exp_H(1\cdot d\varphi(X)).
\end{align*}
Since scalar multiplication by $1$ is the identity on the vector spaces $\mathfrak g$ and $\mathfrak h$, this is exactly
\begin{align*}
\varphi(\exp_G X)=\exp_H(d\varphi(X)).
\end{align*}
This completes the proof.
[/step]
