= Solution
Because the <exponential function> is strictly increasing, the <level sets> $f=L>0$ are exactly those of $G=\ln L$. For $a=1$ they satisfy
$$
y^2=2\ln L-x^2-\frac{x^4}{2}.
$$
There are no contours for $L<1$, the level $L=1$ is the origin, and $L>1$ gives nested closed ovals symmetric about both axes. Their stronger fourth-power growth in $x$ makes them relatively narrower horizontally as the level rises.
For $a=-1$, the contour equation becomes
$$
y^2=2\ln L-x^2+\frac{x^4}{2}.
$$
The central <local minimum> is surrounded by closed ovals for $1<L<e^{1/4}$, accompanied by outer open branches. At $L=e^{1/4}$, the level passes through the two <saddle points> $(\pm1,0)$ and changes its connectivity. Levels $L<1$ have only outer branches, while levels above the saddle value have an upper and a lower branch. The sketch is restricted to the requested square; labels give $G=\ln f$, so they identify the corresponding contours of $f$ without exponentially large labels.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1-stationary-contours.png]
{title=Contours for the two parameter signs; labels are log f and dots mark stationary points}
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