 Solution

Write $A(s)=s(s^2-4)$, so $z=A(x)A(y)$ andSolving these simultaneously gives thirteen stationary points: the nine grid points with $x,y\in \{-2,0,2\}$ and the four points $x,y\in \{-2/\sqrt{3},2/\sqrt{3}\}$. There are no mixed extra cases, because $A$ and $A^{\prime }$ have no common root.

At a grid point the Hessian matrix has zero diagonal and off-diagonal entry $A^{\prime }(x)A^{\prime }(y)\ne 0$, so its determinant is negative: all nine are saddle points. At the other four points the Hessian matrix is $-16xyI$. Thus the equal-sign points are local maxima with value $256/27$, and the opposite-sign points are local minima with value $-256/27$.

In the requested nonnegative quadrant, the stationary points are the four boundary saddle points $(0,0),(0,2),(2,0),(2,2)$ and the interior local maximum $(2/\sqrt{3},2/\sqrt{3})$. The zero contours are $x=0,2$ and $y=0,2$. The gradient is perpendicular to regular contours and points toward increasing $z$. A boundary point's classification above refers to the function on the whole plane; its behavior restricted to the quadrant can be one-sided. The interior local maximum is not a global maximum on the unbounded quadrant, since $z\to +\infty$ along $x=y\to \infty$.