The left-hand side decomposes as $$\left(x^{6}+3 x^{4} y^{2}+3 x^{2} y^{4}+y^{6}\right)-\left(x^{5}-10 x^{3} y^{2}+5 x y^{4}\right)=\left(x^{2}+y^{2}\right)^{3}-\left(x^{5}-10 x^{3} y^{2}+5 x y^{4}\right)$$. Now, note that $$(x+i y)^{5}=x^{5}+5 i x^{4} y-10 x^{3} y^{2}-10 i x^{2} y^{3}+5 x y^{4}+i y^{5}$$ so that our function is just $\left(x^{2}+y^{2}\right)^{3}-\Re\left((x+i y)^{5}\right)$. Switching to polar coordinates, this is $r^{6}-\Re\left(r^{5}(\cos \theta+i \sin \theta)^{5}\right)=r^{6}-r^{5} \cos 5 \theta$ by de Moivre's rule. The graph of our function is then the graph of $r^{6}-r^{5} \cos 5 \theta=0$, or, more suitably, of $r=\cos 5 \theta$. This is a five-petal rose, so the answer is 5.