Answer
For $\frac{x^2}{x+y}=y^2+1$ find $\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{x^2+2xy}{(x+y)^22y+x^2}$
Work Step by Step
Differentiate both sides with respect to x
$\frac{2x(x+y)-x^2(1+\frac{dy}{dx})}{(x+y)^2}=2y\frac{dy}{dx}$
Isolate $\frac{dy}{dx}$
$2x^2+2xy-x^2-x^2\frac{dy}{dx}=\frac{dy}{dx}(x+y)^22y$
$x^2+2xy=\frac{dy}{dx}(x+y)^22y+x^2\frac{dy}{dx}$
$x^2+2xy=\frac{dy}{dx}((x+y)^22y+x^2)$
$\frac{dy}{dx}=\frac{x^2+2xy}{(x+y)^22y+x^2}$