Polar contour plot in Mathematica?

You can use TransformedField to get a function that can be used as the first argument of ContourPlot:

 f = (r^2 - a^3/r) Sin[t]^2;
 tf = TransformedField[ "Polar" -> "Cartesian", f, {r, t} -> {x, y}]

TeXForm @ tf

$\frac{y^2 \left(x^2 \sqrt{x^2+y^2}+y^2 \sqrt{x^2+y^2}-1\right)}{\left(x^2+y^2\right)^{3/2}}$

cValues = {0.00001, 0.01, 0.05, 0.1, 0.3, 0.6, 1.0, 1.5, 2.0, 2.5, 3.2};
a = 1;

ContourPlot[tf, {x, -3, 3}, {y, -3, 3}, 
  Contours -> cValues, 
  PlotPoints-> 200,
  Axes -> True,
  Frame -> False,
  PlotRange -> All, 
  ContourShading -> None, 
  AspectRatio -> Automatic,
  RegionFunction -> (Norm[{#, #2}] <= 3&)]

An alternative approach is to use f with ContourPlot and post-process the output to transform the lines:

cp1 = ContourPlot[f, {r, 0, 3}, {t, -Pi, Pi}, 
       Contours -> cValues, PlotRange -> All, 
       ContourShading -> None,  Axes -> True, 
       Frame -> False, ImageSize -> 300];

cp2 = Show[cp1 /. GraphicsComplex[c_, rest___] :> 
        GraphicsComplex[c /. {a_, b_} :> (a {Cos[b], Sin[b]}), rest], 
    AspectRatio -> Automatic, ImageSize -> 300];

Row[{cp, cp2}, Spacer[15]]

Here is how to do the coordinate system conversion by hand:

cValues = {0.00001, 0.01, 0.05, 0.1, 0.3, 0.6, 1.0, 1.5, 2.0, 2.5, 
   3.2};

ContourPlot[
 (Norm[{x, y}]^2 - 3/Norm[{x, y}]) Sin[ArcTan[x, y]]^2,
 {x, -3, 3},
 {y, -3, 3},
 Contours -> cValues
 ]

Using MeshFunctions and Mesh in a ParametricPlot of polar coordinates to define the contours:

cValues = {0.00001, 0.01, 0.05, 0.1, 0.3, 0.6, 1.0, 1.5, 2.0, 2.5, 3.2};
Block[{a = 1},
 ParametricPlot[r {Cos[\[Theta]], Sin[\[Theta]]},
  {r, 0, 3 a}, {\[Theta], 0, 2 Pi},
  PlotStyle -> None, BoundaryStyle -> None, PlotPoints -> {60, 120},
  MeshFunctions ->
   {Function[{x, y, r, \[Theta]}, (r^2 - a^3/r) Sin[\[Theta]]^2]}, 
  Mesh -> {cValues}, 
  MeshStyle -> {Directive[ColorData[97][1], AbsoluteThickness[1.6]]}, 
  PlotRange -> {All, {-2, 2}}, Method -> {"BoundaryOffset" -> True}]
 ]