Answer
$$\frac{dy}{dx}=\frac{1}{y(x+1)^2}$$
Work Step by Step
$$y^2=\frac{x-1}{x+1}$$
1) Differentiate both sides of the equation with respect to $x$:
$$\frac{d}{dx}(y^2)=\frac{d}{dx}\Big(\frac{x-1}{x+1}\Big)$$ $$2y\frac{dy}{dx}=\frac{(x+1)-(x-1)}{(x+1)^2}$$ $$2y\frac{dy}{dx}=\frac{x+1-x+1}{(x+1)^2}$$ $$2y\frac{dy}{dx}=\frac{2}{(x+1)^2}$$ $$y\frac{dy}{dx}=\frac{1}{(x+1)^2}$$
2) Now solve for $dy/dx$: $$\frac{dy}{dx}=\frac{1}{y(x+1)^2}$$
