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.)
 A: Allowing $r$ to take both nonnegative and negative values, the graphs of $r=\sin2\theta$ and $r=-\sin2\theta$ (on $0\le\theta<2\pi$) are identical—they contain the same set of geometric points in the plane—though they are traced out differently.  For $r=\sin2\theta$, as $\theta$ increases from $0$ to $2\pi$, the graph traces the first quadrant petal, then the fourth quadrant petal, then the third quadrant petal, and finally the second quadrant petal.  For $r=-\sin2\theta$, as $\theta$ increases from $0$ to $2\pi$, the graph traces the third quadrant petal, then the second quadrant petal, then the first quadrant petal, and finally the fourth quadrant petal.
edit Here's what I mean:

Since the sets of points in the plane described by $r=\sin2\theta$ and $r=-\sin2\theta$ are identical, giving only one of them is sufficient to describe the same set of points as $(x^2+y^2)^3=4x^2y^2$.  That's what lets us go from $r^2=\sin^22\theta$ and its equivalent of $r=\pm\sin2\theta$ to only $r=\sin2\theta$.
A: If $x=r\cos(\theta)$ and $y=r\sin(\theta)$, then $(x^2+y^2)^3=4x^2y^2$ becomes
$$
r^2=\sin^2(2\theta)
$$
as you say. However, note that the curve $r=\sin(2\theta)$ is symmetric through the origin since $\sin(2(\theta+\pi))=\sin(2\theta)$, so it is the same as $r=-\sin(2\theta)$. In other words, it doesn't matter whether you take $+$ or $-$.
$\hskip{4cm}$ 
A: On the variety $V\left((x^2+y^2)^3-(2xy)^2\right)$,
in polar coordinates we have
$$
r^6=(x^2+y^2)^3=(2xy)^2=r^4\sin^2{2\theta}
$$
or
$$
r^2=\sin^2{2\theta}
\quad\implies\quad
r=\pm\sin2\theta
$$
(we can ignore the "trivial" algebraic solution $r=0$ for all $\theta$
since it is geometrically recovered for $\theta\in\pi\mathbb{Z}$).
Now $\sin{2\theta}$ has period $\pi$
(furnishing the same magnitude for $r$ at antipodal points),
the curve has point symmetry through the origin,
allowing us to discard the sign and consider only $r\in[0,1]$.
But
$$
\sin^2{2\theta}=\frac{1-\cos{4\theta}}{2}
$$
has period $\frac{\pi}{2}$, vanishes at the axes or "cardinal" directions,
and has its maxima of $1$ when $\cos{4\pi}=-1$, i.e. at $\theta=\frac{\pi}{4}+k\frac{\pi}{2}$.
This explains why the polar equation for the solution can take the forms
$$
r=\left|\sin{2\theta}\right|
=\left(\sin^2{2\theta}\right)^\frac{1}{2}
=\left(\frac{1-\cos{4\theta}}{2}\right)^\frac{1}{2}
$$
and why it gives us a four-leaved rose.
Qualitatively, the result is the same
with any positive power $r=|\sin{2\theta}|^p$,
but the petal thickness and area and
the attenuation of the maximum radius
all vary inversely with $p$.
Plotted with sage (online), here is
the solution $r^2=\sin^2{2\theta}$ (with the power $p=1$) in red,
with another curve $r^6=\sin^2{2\theta}$ (with $p=\frac{1}{3}$) in blue
to illustrate the effect of $p$.
t=var('t')
(polar_plot(abs(sin(2*t))^(2/2), (t, 0, 2*pi), color='red')
+polar_plot(abs(sin(2*t))^(2/6), (t, 0, 2*pi), color='blue')).show()


Finally, as a refresher, a good way to remember or derive
the angle-doubling step above is from the two equations
$$
\begin{matrix}
\cos^2t&+&\sin^2t&=&1\\
\cos^2t&-&\sin^2t&=&\cos2t.
\end{matrix}
$$
Adding or subtracting (and dividing by two) then yields
$$
\cos^2t = \frac{1+\cos2t}{2}
\qquad
\text{or}
\qquad
\sin^2t = \frac{1-\cos2t}{2}
$$
respectively.
A: Avoiding the issue entirely: 
$$r = \sin(2\theta) = 2\sin\theta\cdot \cos\theta$$ 
$$r^3 = 2(r\sin\theta)(r\cos\theta)$$
$$x = r\cos\theta$$
$$y = r\sin\theta$$ 
$$r^3 =2xy$$
$$r = (x^2 + y^2)^{\frac 12}$$
$$(x^2+y^2)^{\frac 32} =2xy$$
$$(x^2+y^2)^3=4x^2y^2$$
