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.