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I have a problem with making contour plots of the form $f(x,y,c)=0$ work well with Mathematica, where c defines the levelset. For example the lemniscate $(x^2 + y^2)^2 - c (x^2 - y^2) =0$. Obviously rewriting this as $frac(x^2 + y^2)^2(x^2 - y^2) =c$ is going to cause some problems for anywhere $x^2=y^2$. If I try to plot multiple curves

ContourPlot[Evaluate@Table[(x^2 + y^2)^2 - m (x^2 - y^2) == 0, m, 0,4],
x, -2, 2, y, -2, 2]

this works but I lose the nice shading in between curves and hover over information about the level sets as seen in this example:

ContourPlot[x^2 + y^2, x, -2, 2, y, -2, 2]

The best I could do is the following:

lemniscates = 
 Table[
 RegionPlot[(x^2 + y^2)^2 - m (x^2 - y^2) < 0, x, -2, 
 2, y, -2, 2, PerformanceGoal -> "Quality", 
 BoundaryStyle -> None, 
 PlotStyle -> ColorData["RedBlueTones", 0, 7][m]], m, 0, 
 5]
Show[Reverse@lemniscates, 
 Prolog -> ColorData["RedBlueTones", 0, 7][7], 
 Rectangle[Scaled[0, 0], Scaled[1, 1]]]

but this looks awful compared to the default contour plotting and I still lose the level set information. How do I get my lemniscate plots to look like the ContourPlot of the circles example?

Thank you!

EDIT: It seems I used a bit too simple an example with the lemniscates to illustrate my problem. Here is another similar function:

ContourPlot[
 Evaluate@Table[(x^2 + m y^2)^2 - m (x^2 - y^2) == 0, m, 0, 
 4], x, -2, 2, y, -2, 2]

But now solving for m gives you two roots so this will not be easy to put in the required form. What can be done?

Generate separate ContourPlot for each m with contour 0 and change the tooltip label to m and combine all with Show:

Show @ Table[ContourPlot[Evaluate[(x^2 + y^2)^2 - m (x^2 - y^2)], x, -2, 2, y, -2, 2, 
 ContourStyle -> Directive[Thick, ColorData[63][m]], 
 ContourShading -> Opacity[.2, ColorData[63][m]], White, 
 Contours -> 0] /. Tooltip[a_, _] :> Tooltip[a, m], m, 0, 4]

Generate the plots in the opposite order to get opaque colors without blending:

colors = Cyan, Red, Blue, Orange, Green;
Show @ Table[ContourPlot[(x^2 + y^2)^2 - m (x^2 - y^2) , x, -2, 2, y, -2, 2, 
 ContourStyle -> Directive[Thick, Darker@Darker@colors[[m + 1]]], 
 ContourShading -> Opacity[1, colors[[m + 1]]], None, 
 Contours -> 0] /. Tooltip[a_, _] :> Tooltip[a, m], 
 m, 4, 0, -1]

If you can tolerate solving for $1/m$ rather than $m$ (as $m=0$ doesn't given you real numbers), you could use the following:

ContourPlot[(x^2 - y^2)/(x^2 + y^2)^2, x, -2, 2, y, -2, 2,
 Contours -> 1, 1/2, 1/3, 1/4, PlotRange -> All,
 ContourShading -> White, Red, Blue, Green, Cyan]

Maybe a more interesting display might be with ContourPlot3D which also allows your actual form of $f(x,y,c)=0$:

ContourPlot3D[(x^2 + y^2)^2 - m (x^2 - y^2) == 0, x, -2, 2, y, -2, 2, m, 0, 4, 
 PlotRange -> All, PlotPoints -> 100]

To get ContourPlot to do what you seek you need to solve for the parameter in the form $m = F(x,y)$; then plot $F(x,y)$ with the desired contours $m$. However, in this OP's case, this causes some numerical problems, which can be alleviated with the following dodge:

ContourPlot[
 (x^2 + y^2)^2/($MachineEpsilon + Abs[x^2 - y^2]),
 x, -2, 2, y, -2, 2,
 Contours -> Table[m, m, 0, 4], 
 RegionFunction -> 
 Function[x, y, z, x^2 != y^2 && 0 <= (x^2 + y^2)^2/(x^2 - y^2) < 5], 
 MaxRecursion -> 3, PlotPoints -> 25]