if tan(x^2-y^2/x^2+y^2)=a then prove that dy/dx=y/x

the given equation is:
$$\begin{gathered} \tan \left(\frac{x^2-y^2}{x^2+y^2}\right)=a \\ \frac{x^2-y^2}{x^2+y^2}={\tan }^{-1}a \\ \frac{x^2+y^2-2y^2}{x^2+y^2}={\tan }^{-1}a \\ 1-2.\frac{y^2}{x^2+y^2}={\tan }^{-1}a \end{gathered}$$
now differentiating on the both sides wrt x:
$$\begin{gathered} 0-2.\left[\frac{(x^2+y^2).2y{\displaystyle \frac{dy}{dx}}-y^2.(2x+2y.{\displaystyle \frac{dy}{dx}})}{(x^2+y^2{)}^2}\right]=0 \\ \Rightarrow \frac{(x^2+y^2).2y{\displaystyle \frac{dy}{dx}}-y^2.(2x+2y.{\displaystyle \frac{dy}{dx}})}{(x^2+y^2{)}^2}=0 \\ \Rightarrow (x^2+y^2).2y\frac{dy}{dx}-y^2.(2x+2y.\frac{dy}{dx})=0 \\ \Rightarrow \left(x^2+y^2-y^2\right).2y\frac{dy}{dx}=2x{y}^2 \\ \Rightarrow 2x^{2}y\frac{dy}{dx}=2x{y}^2 \\ \frac{dy}{dx}=\frac{y}{x} \end{gathered}$$
which is the required result.

hope this helps you