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Added in ContourPlot3D for more complicated example that the OP recently added.
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JimB
JimB
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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}]

Contours

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]

ContourPlot3D example

For your more complex example, I think it becomes more important to consider ContourPlot3D:

ContourPlot3D for more complicated example

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}]

Contours

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]

ContourPlot3D example

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}]

Contours

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]

ContourPlot3D example

For your more complex example, I think it becomes more important to consider ContourPlot3D:

ContourPlot3D for more complicated example

Added ContourPlot3D example.
Source Link
JimB
JimB
  • 42.9k
  • 3
  • 51
  • 108

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}]

Contours

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},
 Contours -> 
 {0}, PlotRange -> All, PlotPoints -> 100]

ContourPlot3D example

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}]

Contours

Maybe a more interesting display might be with ContourPlot3D:

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

ContourPlot3D example

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}]

Contours

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]

ContourPlot3D example

Added ContourPlot3D example.
Source Link
JimB
JimB
  • 42.9k
  • 3
  • 51
  • 108

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}]

Contours

Maybe a more interesting display might be with ContourPlot3D:

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

ContourPlot3D example

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}]

Contours

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}]

Contours

Maybe a more interesting display might be with ContourPlot3D:

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

ContourPlot3D example

Source Link
JimB
JimB
  • 42.9k
  • 3
  • 51
  • 108