Writing a Polar Equation for the Graph of an Implicit Cartesian Equation 
If $(x^2+y^2)^3=4x^2y^2,$ then $r=\sin 2\theta$ for some $\theta$.

Using $r^2=x^2+y^2, x=r\cos\theta,y=r\sin\theta$, it's easy to get $r^2=\sin^22\theta$.
But I don't know what to do next, since $r$ could be negative in $r=\sin2\theta.$
Actually  the original problem is to show that the affine variety $V((x^2+y^2)^3-4x^2y^2)$ is contained in the four-leaved rose, whose polar equation is certainly $r=\sin 2\theta$. (Exercise 7(b), section 1.2, Ideals, Varieties and Algorithms, 3rd edition, David Cox etc.)

Avoiding the issue entirely: 
$$r = \sin(2\theta) = 2\sin\theta\cdot \cos\theta$$ 
$$r^3 = 2(r\sin\theta)(r\cos\theta)$$
$$x = r\cos\theta$$
$$y = r\sin\theta$$ 
$$r^3 =2xy$$
$$r = (x^2 + y^2)^{\frac 12}$$
$$(x^2+y^2)^{\frac 32} =2xy$$
$$(x^2+y^2)^3=4x^2y^2$$
