= Solution
In the translated coordinates $u=x-1$, $v=y+1$, the zero <level set> is the pair of <lines> $u=0$ and $v=0$. The <function> is positive in the quadrants $uv>0$ and negative in those with $uv<0$, and tends to zero as $u^2+v^2\to\infty$. Thus its nonzero <contour plot>[contours] form closed curves around the four off-centre <stationary points>.
For an explicit description of the <contour plot>, use <polar coordinates> $u=r\cos\theta$, $v=r\sin\theta$:
$$
f=\frac{r^2}{2}e^{-r^2/2}\sin2\theta.
$$
At fixed direction within a quadrant, the magnitude increases until $r=\sqrt2$ and then decreases towards zero. Also $|u|e^{-u^2/2}\leq e^{-1/2}$, with equality exactly when $|u|=1$, and similarly for $v$. Consequently $|f|\leq e^{-1}$; each positive level below $e^{-1}$ encloses a positive peak in each positive quadrant, with the analogous negative contours in the negative quadrants.
To classify the <stationary points>, the <Hessian matrix> is
$$
\nabla^2f=E\begin{pmatrix}uv(u^2-3)&(1-u^2)(1-v^2)\\(1-u^2)(1-v^2)&uv(v^2-3)\end{pmatrix}.
$$
At $u=v=0$ it is $\begin{pmatrix}0&1\\1&0\end{pmatrix}$, with <eigenvalues> $1$ and $-1$, so $(1,-1)$ is a <saddle point>. At $u,v\in\{-1,1\}$ it is $-2fI$. Therefore \b[$(2,0)$ and $(0,-2)$ are strict <local maxima>, and $(2,-2)$ and $(0,0)$ are strict <local minima>]. These are also global maxima and minima by the bound just proved.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1-q16-contours.png]
{title=Contours and gradient directions near all five stationary points in Question 16}
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