You could use Lagrange multipliers to maximize $f(x,y)=y$ subject to the constraint that$$g(x,y) = x^2 + y^2 - (2 x^2 + 2 y^2 - x)^2 = 0.$$
f[x_, y_] = y;g[x_, y_] = x^2 + y^2 - (2 x^2 + 2 y^2 - x)^2;eqs = {D[f[x, y], x] == lambda*D[g[x, y], x], D[f[x, y], y] == lambda*D[g[x, y], y], g[x, y] == 0};Solve[eqs, {x, y, lambda}] // InputForm(* Out: { {x -> 3/8, y -> (-3*Sqrt[3])/8, lambda -> 2/(3*Sqrt[3])}, {x -> 3/8, y -> (3*Sqrt[3])/8, lambda -> -2/(3*Sqrt[3])}}*)The maximum value of $y$ is $3\sqrt{3}/8 \approx 0.6495$. Of course, this should occur where the proper contour of $y$ is tangent to the restraint curve. You can visualize the situation like so.
contourPic = ContourPlot[y, {x, -1, 2}, {y, -1, 1}, Contours -> Range[-2, 2, 1/2]*3 Sqrt[3]/8];restraintPic = ContourPlot[x^2 + y^2 - (2 x^2 + 2 y^2 - x)^2 == 0, {x, -1, 2}, {y, -1, 1}, ContourStyle -> {Thick, Black}];Show[{contourPic, restraintPic}, AspectRatio -> Automatic, Epilog -> {PointSize[Large], Blue, Point[{3/8, 3 Sqrt[3]/8}]}]Image may be NSFW.
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