= Solution
In the $(x,y)$-plane take $z=0$; holding another fixed $z$ merely shifts all contour labels by $z^2$. The <level sets> satisfy
$$
x^2=y^4-2ay^2+c.
$$
For $a=1$, the origin is a <local minimum> and $(0,\pm1)$ are <saddle points of a scalar function> at contour value $1$. For $0<c<1$, a closed contour around the origin is accompanied by unbounded outer components. At $c=1$ the contour satisfies $x=\pm(y^2-1)$ and crosses at the two <saddle points of a scalar function>. For $c>1$, two unbounded components remain; negative contours have only outer components.
For $a=-1$, the origin is a <saddle point of a scalar function>, and the zero contour is $x=\pm y\sqrt{y^2+2}$, with tangent lines $x=\pm\sqrt2\,y$. Positive contours form two components extending on opposite sides of the $y$-axis; negative contours form upper and lower components. These <level sets> are illustrated below.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-2-level-contours.png]
{title=Contours of the quartic surface at z equal to zero for a equal to one and minus one}
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