= Solution
The <partial derivatives> give $x^3=9y$ and $y^3=9x$. If either coordinate is zero, both vanish. Otherwise substitution gives $x^8=3^8$, so the real <stationary points> are $(0,0)$, $(3,3)$ and $(-3,-3)$.
The <Hessian matrix> is
$$
H_g=\begin{pmatrix}12x^2&-36\\-36&12y^2\end{pmatrix}.
$$
At the origin its <eigenvalues> are $36$ and $-36$, so it is a saddle at level zero. At either of the other points its <eigenvalues> are $72$ and $144$, so both are strict minima at level $-162$. In fact,
$$
g=(x^2-y^2)^2+2(xy-9)^2-162\geq-162,
$$
with equality only at those two points. Hence
$$
\boxed{(0,0):\text{ saddle, }g=0;\qquad
(3,3),(-3,-3):\text{ global minima, }g=-162.}
$$
For levels $-162<c<0$, two separate closed contours surround the two minima. At $c=0$ the contours form two lobes, in the first and third quadrants, meeting at the origin. Their limiting tangent directions there are the coordinate axes because the leading term is $-36xy$. On the diagonal the outer zero-level points are $(\pm3\sqrt2,\pm3\sqrt2)$. Positive levels form one closed outer contour enclosing both wells; there are no contours below $-162$. The graph is symmetric under exchanging $x,y$ and under reversing both signs, but not under a quarter-turn. These features are the $a=3$ case of the <quartic double well with bilinear coupling>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1-quartic-contours.png]
{title=The quartic function's two minima, saddle, and transition from two contour loops to one}
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