Answer
$y'=\dfrac{x^{2}+2xy}{x^{2}+2y(x+y)^{2}}$
Work Step by Step
$\dfrac{x^{2}}{x+y}=y^{2}+1$
Differentiate the whole equation:
$\Big(\dfrac{x^{2}}{x+y}\Big)'=(y^{2}+1)'$
Apply the quotient rule to differentiate the left side:
$\dfrac{(x+y)(x^{2})'-(x^{2})(x+y)'}{(x+y)^{2}}=(2y)(y')$
$\dfrac{(x+y)(2x)-(x^{2})(1+y')}{(x+y)^{2}}=2yy'$
Solve for $y'$:
$2x^{2}+2xy-x^{2}-x^{2}y'=2yy'(x+y)^{2}$
$-x^{2}y'-2yy'(x+y)^{2}=-x^{2}-2xy$
$y'[-x^{2}-2y(x+y)^{2}]=-x^{2}-2xy$
$y'=\dfrac{-x^{2}-2xy}{-x^{2}-2y(x+y)^{2}}$
$y'=\dfrac{x^{2}+2xy}{x^{2}+2y(x+y)^{2}}$
