= Solution
Let $G=x^3e^{-x^2-y^2}$ and $a=\sqrt{3/2}$. Its <partial derivatives> are
$$
G_x=x^2(3-2x^2)e^{-x^2-y^2},\qquad G_y=-2yx^3e^{-x^2-y^2},
$$
so both vanish at $(a,0)$ and $(-a,0)$. Evaluating the second derivatives at a point $(x,0)$ with $x^2=3/2$ gives
$$
G_{xx}=-6xe^{-3/2},\qquad G_{xy}=0,\qquad G_{yy}=-3xe^{-3/2}.
$$
Therefore the <Hessian determinant> is $18x^2e^{-3}=27e^{-3}>0$. Its two diagonal entries are negative at $x=a$ and positive at $x=-a$. Hence
$$
\boxed{\begin{aligned}
(a,0)&:\ \text{strict local maximum},&G&=\left(\frac{3}{2e}\right)^{3/2},\\
(-a,0)&:\ \text{strict local minimum},&G&=-\left(\frac{3}{2e}\right)^{3/2}.
\end{aligned}}
$$
These are also global extrema: for positive $x$, the factor $e^{-y^2}\le1$ maximizes the positive value at $y=0$, and differentiating $x^3e^{-x^2}$ fixes its positive maximum at $a$. Oddness in $x$ supplies the corresponding global minimum.
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