Find the values

We are given with an equation $y^{3}-3 x y^{2}=x^{3}+3 x^{2} y$, we have to find $\frac{d y}{d x}$ of it, so by differentiating the equation on both sides with respect to $x$, we get,

$3 y^{2} \frac{d y}{d x}-3\left[y^{2}(1)+2 x y \frac{d y}{d x}\right]=3 x^{2}+3\left[2 x y+x^{2} \frac{d y}{d x}\right]$

Taking $\frac{d y}{d x}$ terms to left hand side and taking common $\frac{d y}{d x}$, we get,

$\frac{d y}{d x}\left[3 y^{2}-6 x y-3 x^{2}\right]=3 x^{2}+6 x y+3 y^{2}$

$\frac{d y}{d x}=\frac{3 x^{2}+3 y^{2}+6 x y}{3 y^{2}-3 x^{2}-6 x y}=\frac{x^{2}+y^{2}+2 x y}{y^{2}-x^{2}-2 x y}$