= Solution
The most useful guide to the <contour plot> is the common saddle value. Direct <polynomial factorization> gives
$$
6f-1=(2y+1)\bigl[3x^2-(y-1)^2\bigr].
$$
Thus the <level set> $f=1/6$ is exactly the three complete lines
$$
\boxed{y=-\frac12,\qquad y=1+\sqrt3x,\qquad y=1-\sqrt3x}.
$$
Their intersections are the three <saddle points>, and they enclose the origin in an equilateral triangle. Near the <local minimum>, the <level sets> are nearly circles because the leading term is $(x^2+y^2)/2$. For $0<c<1/6$, a closed contour surrounds the origin, while three separate open branches lie outside the triangular barrier. At the saddle level these components meet at the <saddle points>; above it the central closed contour has opened into the three outer channels.
The identity in <polar coordinates>
$$
f=\frac12r^2+\frac13r^3\sin3\theta
$$
shows the rotation symmetry through $2\pi/3$. It identifies the function as the <Hénon-Heiles potential> and helps organize all three channels in the original sketch below. The dashed straight lines are the saddle level; labels on the other contours give $f$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2-q12-contours.png]
{title=Contours and gradient directions of the Hénon-Heiles potential}
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