Answer
$(x-y)^2+\frac{y}{x}$
Work Step by Step
$$x^2=\frac{x+y}{x-y}$$
$$\frac{d}{dx}x^2=\frac{d}{dx}\frac{x+y}{x-y}$$
$$2x=\frac{d}{dx}\frac{x+y}{x-y}$$
According to the quotient rule,
$$2x=\frac{(x-y)\frac{d}{dx}(x+y)-(x+y)\frac{d}{dx}(x-y)}{(x-y)^2}$$
$$2x=\frac{(x-y)(1+\frac{dy}{dx})-(x+y)(1-\frac{dy}{dx})}{(x-y)^2}$$
Simplifying,
$$2x=\frac{(x-y+x\frac{dy}{dx}-y\frac{dy}{dx})-(x+y-x\frac{dy}{dx}-y\frac{dy}{dx})}{(x-y)^2}$$
$$2x=\frac{-2y+2x\frac{dy}{dx}}{(x-y)^2}$$
Solving for $\frac{dy}{dx}$,
$$2x(x-y)^2=-2y+2x\frac{dy}{dx}$$
$$2x(x-y)^2+2y=2x\frac{dy}{dx}$$
$$\frac{2x(x-y)^2+2y}{2x}=\frac{dy}{dx}$$
Simplifying,
$$\frac{x(x-y)^2+y}{x}=\frac{dy}{dx}$$
$$\frac{dy}{dx}=(x-y)^2+\frac{y}{x}$$
