$\begin{aligned} & y=x \tan y \\ & \therefore \quad \frac{d y}{d x}=x \sec ^2 y \frac{d y}{d x}+\tan y \\ & \therefore \quad\left(x \sec ^2 y-1\right) \frac{d y}{d x}=-\tan y \\ & \therefore \quad \frac{d y}{d x}=\frac{-\tan y}{x \sec ^2 y-1}=\frac{-x \tan y}{x^2 \sec ^2 y-x} \\ & =\frac{-x \tan y}{x^2\left(1+\tan ^2 y\right)-x}=\frac{-x \tan y}{x^2+x^2 \tan ^2 y-x} \\ & \therefore \quad \frac{d y}{d x}=\frac{-y}{x^2+y^2-x}=\frac{y}{x-x^2-y^2} \cdot \cdots[\because y=x \tan y, \text { given }]\end{aligned}$