 Solution

Let $G=x^3{e}^{-x^2-y^2}$ and $a=\sqrt{3/2}$. Its partial derivatives areso both vanish at $(a,0)$ and $(-a,0)$. Evaluating the second derivatives at a point $(x,0)$ with $x^2=3/2$ givesTherefore the Hessian determinant is $18x^2{e}^{-3}=27e^{-3}>0$. Its two diagonal entries are negative at $x=a$ and positive at $x=-a$. HenceThese are also global extrema: for positive $x$, the factor $e^{-y^2}\le 1$ maximizes the positive value at $y=0$, and differentiating $x^3{e}^{-x^2}$ fixes its positive maximum at $a$. Oddness in $x$ supplies the corresponding global minimum.