= Solution
The <gradient> vanishes when $3x^2-2=3y^2-2=0$. Put $s=\sqrt{2/3}$. The <Hessian matrix> is $\operatorname{diag}(6x,6y)$, so the <second-derivative test> gives:
* $(s,s)$: a \b[strict local minimum], with $f=-8s/3$;
* $(-s,-s)$: a \b[strict local maximum], with $f=8s/3$;
* $(s,-s)$ and $(-s,s)$: \b[saddle points], both with $f=0$.
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,
$$
f=(x+y)(x^2-xy+y^2-2),
$$
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.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016-stationary-contours.png]
{title=Contours of the cubic surface, with its two local extrema, two saddles and exact zero contour}
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