= Solution
For $c\ne0$, <completing the square> gives the <conic section>
$$
(c+2)x^2+c\left(y-\frac1c\right)^2=\frac{1-c^2}{c}.
$$
The possible levels follow from
$$
f+2=\frac{2(y^2+y+1)}{D}>0,\qquad f\leq1:
\qquad \boxed{-2<f\leq1.}
$$
For $0<c<1$, the contours are <ellipses> centred at $(0,1/c)$, shrinking to the maximum as $c\uparrow1$. The zero contour is the <parabola> from part (c). For $-1<c<0$, the contours are vertical-opening <hyperbolas>. At $c=-1$ they degenerate into the intersecting straight lines, and for $-2<c<-1$ they are horizontal-opening <hyperbolas>. These descriptions specify the remaining contour families and their symmetry about the $y$ axis. The lower bound $-2$ is approached as $|x|\to\infty$ at fixed $y$ but is never attained.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2-q12-gradient-contours.png]
{title=Contours and an ascending <gradient> trajectory}
{description=Original contours of the rational scalar field. The critical points are marked; the green parabola is the zero contour and the crossing dashed lines have level $-1$. The red trajectory starts at $(1,1)$ and follows the positive <gradient> towards the maximum, as discussed in part (e).}
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