# Plotting $(x^2 + y^2 - 1)^3 - x^2 y^3 = 0$

I have no idea how this equation: $$(x^2 + y^2 - 1)^3 - x^2 y^3 = 0$$

Produces this picture:

Can someone provide a general explanation of plotting this function?

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The answer depends on what kind of software you wish to use. The picture you provide looks as if it has been generated by Mathematica. The command would be ContourPlot[f[x,y],{x,-2,2},{y,-2,2}]. – Eckhard Dec 22 '12 at 14:54
@Eckhard: I'm asking of some explanation like this. – m0nhawk Dec 22 '12 at 15:01
In general, plotting implicit functions is done by algorithms such as marching squares. – Peter Sheldrick Dec 22 '12 at 19:00

The solution set is obviously symmetric with respect to the $y$-axis. Therefore we may assume $x\geq 0$. In the domain $\{(x,y)\in {\mathbb R}^2\ |\ x\geq0\}$ the equation is equivalent with $$x^2+ y^2 -1=x^{2/3} y\ ,$$ which can easily be solved for $y$: $$y={1\over2}\bigl(x^{2/3}\pm\sqrt{x^{4/3}+4(1-x^2)}\bigr)\ .$$ Now plot this, taking both branches of the square root into account. You might have to numerically solve the equation $x^{4/3}+4(1-x^2)=0$ in order to get the exact $x$-interval.