(y^3-2x^2y)dx+(2xy^2-x^3)dy=0 represent the curve:

Put y = tx; $x\frac{dt}{dx}=\frac{3t(1-t)(1+t)}{2t^2-1};\int(\frac{1}{t}+\frac{1/2}{t-1}+\frac{1/2}{t+1-1})dt+\int\frac{3}{x}dx=const.$

In $t\sqrt{t^2-1}+3\,ln\,x=const.$

$xy\sqrt{y^2-x^2}=c$