= Solution
The <gradient> components are $f_x=2x(1+6x^2)$ and $f_y=2y(1-6y^2)$. Their real simultaneous zeros are $(0,0)$ and $(0,\pm1/\sqrt6)$. The <Hessian matrix> is diagonal:
$$
H=\operatorname{diag}(2+36x^2,2-36y^2).
$$
At the origin it is positive definite; at each other critical point it is $\operatorname{diag}(2,-4)$ and indefinite. Hence
$$
\boxed{(0,0):\text{ strict local minimum},\ f=0;\qquad
(0,\pm1/\sqrt6):\text{ saddle points},\ f=1/12.}
$$
There are no local maxima. The <contours> are symmetric under either coordinate reflection. Small positive levels have a closed component surrounding the minimum; these meet the saddle-level separatrices at level $1/12$. That critical contour has the useful exact form $3x^4+x^2=3(y^2-1/6)^2$. Other components extend toward large $|y|$, where the $-3y^4$ term makes the function unbounded below.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016-resit-stationary-contours.png]
{title=Contours near a minimum and two saddle points, including the critical level one twelfth}
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