 Solution

The gradient vanishes when $3x^2-2=3y^2-2=0$. Put $s=\sqrt{2/3}$. The Hessian matrix is $\mathrm{diag}(6x,6y)$, so the second-derivative test gives:

The positive-definite and negative definite cases give small closed level curves around the minimum and maximum. At each saddle, the zero contour crosses itself. In fact,so the entire zero level curve is the union of the line $y=-x$ and the ellipse $x^2-xy+y^2=2$. This factorization fixes the topology of a rough contour sketch. Neither local extremum is global, since the cubic is unbounded above and below.