= Solution
At $(X,y)=(0,0)$ the <Hessian matrix> is $\begin{pmatrix}0&1\\1&0\end{pmatrix}$, with <eigenvalues> $1$ and $-1$, so $(x,y)=(1,0)$ is a <saddle point>. At the other four points the mixed second <derivative> vanishes and the two diagonal entries both equal $-2Xy/e$. Hence
$$
\boxed{\begin{array}{c|c|c}
(x,y)&\text{type}&z\\\hline
(2,1),(0,-1)&\text{maxima}&e^{-1}\\
(0,1),(2,-1)&\text{minima}&-e^{-1}\\
(1,0)&\text{saddle}&0
\end{array}.}
$$
The four extrema are global: $|z|=(|X|e^{-X^2/2})(|y|e^{-y^2/2})\leq e^{-1}$ because each factor has maximum $e^{-1/2}$ at absolute argument one.
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