= Solution
The objective is <continuous> on the compact disk, so a <global maximum> exists. Its <gradient> is $(e^x,2e^{2y})$, which is never zero, so a maximum cannot lie in the interior. It must have $x^2+y^2=1$.
Both coordinates of a maximizing point are positive. Reflecting a negative coordinate to its positive counterpart strictly increases the objective without changing the constraint; at an endpoint with one coordinate zero, a small move along the circle into the positive quadrant also increases the objective to first order. Thus it suffices to consider $x>0,y>0$.
The boundary <Lagrange multiplier> equations are $e^x=2\lambda x$ and $2e^{2y}=2\lambda y$. Eliminate $\lambda$ to obtain the two-dimensional nonlinear <system of equations>
$$
\boxed{F(x,y)=
\begin{pmatrix}x^2+y^2-1\\ y e^x-2x e^{2y}\end{pmatrix}=0,
\qquad x>0,\ y>0.}
$$
Keeping this quadrant restriction excludes the boundary minimum that an unconstrained elimination could also produce. Under the stated uniqueness assumption for the admissible solution, the <global maximum> must be that solution. Approximate it with <multivariable Newton iteration>, solving $J_F(z_n)\Delta_n=-F(z_n)$ and setting $z_{n+1}=z_n+\Delta_n$, with a suitable safeguard if needed. For example its explicit <Jacobian matrix> is
$$
J_F(x,y)=\begin{pmatrix}
2x&2y\\
y e^x-2e^{2y}&e^x-4x e^{2y}
\end{pmatrix}.
$$
No numerical solution of the system is required.
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