= Solution
Write $A(s)=s(s^2-4)$, so $z=A(x)A(y)$ and
$$
z_x=(3x^2-4)y(y^2-4),\qquad z_y=(3y^2-4)x(x^2-4).
$$
Solving these simultaneously gives \b[thirteen stationary points]: the nine grid points with $x,y\in\{-2,0,2\}$ and the four points $x,y\in\{-2/\sqrt3,2/\sqrt3\}$. There are no mixed extra cases, because $A$ and $A'$ have no common root.
At a grid point the <Hessian matrix> has zero diagonal and off-diagonal entry $A'(x)A'(y)\ne0$, so its <determinant> is negative: all nine are <saddle points>. At the other four points the <Hessian matrix> is $-16xy\,I$. 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 $(2/\sqrt3,2/\sqrt3)$. 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 $x=y\to\infty$.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2022-polynomial-contours.png]
{title=Contours, zeros, stationary points and gradient directions for the polynomial surface in the nonnegative quadrant}
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