 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 satisfyThere 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 becomesThe 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 $(\pm 1,0)$ 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.