Constrained extrema and a compactness argument

Lagrange system.
$$\frac{\partial f}{\partial x} = ay e^{axy}, \quad \frac{\partial g}{\partial x} = 3x^2+1,$$
$$\frac{\partial f}{\partial y} = ax e^{axy}, \quad \frac{\partial g}{\partial y} = 3y^2+1.$$
The condition $\nabla f = \lambda \nabla g$ gives
$$\frac{ay e^{axy}}{3x^2+1} = \frac{ax e^{axy}}{3y^2+1}.$$
Simplifying by $a e^{axy} > 0$:
$$\frac{y}{3x^2+1} = \frac{x}{3y^2+1} \implies y(3y^2+1) = x(3x^2+1).$$
The function $t \mapsto 3t^3+t$ is strictly increasing, so $x = y$.

Critical point. Substituting $x=y$ into the constraint:
$$2x^3 + 2x - 4 = 0 \implies x^3 + x - 2 = 0.$$
The only real root is $x=1$ (since $x^3+x-2 = (x-1)(x^2+x+2)$ and the discriminant of $x^2+x+2$ is negative). So the only possible critical point is $(1,1)$.

Compactness argument. On the curve $G$, if $\|(x,y)\| \to +\infty$, then $x$ and $y$ have opposite signs (since $x^3+y^3$ must stay bounded), so $axy \to -\infty$ and $f(x,y) \to 0$. The function $f$ is continuous on $G$ (which is closed), and $f(1,1) = e^a > 0$. Hence $f$ attains its maximum on a compact set, and this maximum cannot be at infinity (where $f \to 0$). The only candidate is $(1,1)$, so it is the global maximum.