How to find the shape of $x^3 + y^3 - 3xy = 0$ using polar coordinates? There is a curve known as the Folium of Descartes, and it has the equation $x^3 + y^3 - 3axy = 0$. I wanted to know whether it was possible to determine the shape of the case a = 1, using polar coordinates. I have only determined its polar coordinates representation, which is given by
$r = \frac{3 \sin \theta \cos \theta}{\sin^3 \theta + \cos^3 \theta}$
However, I am not too sure how to proceed from here. Do I use calculus or find the values at specific values of $\theta$, and if so, how? Any help would be greatly appreciated. Thank you!

You can write $r^2$ in terms of $\sin 2\theta$, which shows some of the symmetries of the curve: $$(s^3+c^3)^2 = (s^2+c^2)^3 +2s^3c^3 - 3s^2c^2(s^2+c^2)$$so $$r^2=\dfrac{\tfrac{9}{4}\sin^2 2\theta}{1+\tfrac{1}{4}\sin^3 2\theta-\tfrac{3}{4}\sin^2 2\theta}=\dfrac{9\sin^2 2\theta}{4+\sin^3 2\theta-3\sin^2 2\theta}=\dfrac{9\sin^2 2\theta}{(1+\sin 2\theta)(2-\sin 2\theta)^2}.$$ But this curve just isn't very nice in polar co-ordinates: it's much nicer in co-ordinates $X=\dfrac{x+y}{\sqrt{2}}$, $Y=\dfrac{x-y}{\sqrt{2}}$. Then $$3Y^2=\frac{X^2(3-\sqrt{2}X)}{1+\sqrt{2}X}$$ or, if you don't care about scale, (replacing $Y$ with $3Y$, $X$ with $3\sqrt{2}X$), $$Y^2=\frac{2X^2(1-2X)}{1+6X}.$$
