Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 13 / a / ii / Solution

At $(X,y)=(0,0)$ the Hessian matrix is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, 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 $-2Xy/e$. HenceThe four extrema are global: $|z|=(|X|e^{-X^2/2})(|y|e^{-y^2/2})\le e^{-1}$ because each factor has maximum $e^{-1/2}$ at absolute argument one.