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.