= Solution
The <level set> $f=c$ can be rearranged, without changing its domain, as
$$
(c+2)x^2+cy^2-2y+c=0.
$$
The requested special contours are therefore
$$
\boxed{f=0:\ y=x^2;\qquad
f=1:\ (x,y)=(0,1);\qquad
f=-1:\ y+1=\pm x.}
$$
The first is a <parabola>, the second a single point, and the last two intersecting straight lines. Moreover
$$
1-f=\frac{3x^2+(y-1)^2}{D}\geq0,
$$
with equality only at $(0,1)$. Hence $(0,1)$ is a strict global maximum. At $(0,-1)$ the crossing critical contour separates sectors with values above and below $-1$, so it is a <saddle point of a scalar function>.
As a <Hessian matrix> check, $H_f(0,1)=\operatorname{diag}(-3,-1)$ is negative definite, while $H_f(0,-1)=\operatorname{diag}(-1,1)$ is indefinite. There is no stationary minimum.
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