Answer
$\frac{\partial^{2} w}{\partial x^{2}}$ = $\frac{2x^{3}+2y^{2}-6xy-6x^{2}y}{(x^{2}+y)^{3}}$
$\frac{\partial^{2} w}{\partial y^{2}}$ = $\frac{2x^{2}+2x}{(x^{2}+y)^{3}}$
$\frac{\partial^{2} w}{\partial x{\partial y}}$ = $\frac{2x^{3}+3x^{2}-y-2xy}{(x^{2}+y)^{3}}$
Work Step by Step
$\frac{\partial w}{\partial x}$ = $\frac{(x^{2}+y)(1)-(x-y)(2x)}{(x^{2}+y)^{2}}$ = $\frac{-x^{2}+y+2xy}{(x^{2}+y)^{2}}$
$\frac{\partial^{2} w}{\partial x^{2}}$ = $\frac{(x^{2}+y)^{2}(-2x+2y)-(-x^{2}+y+2xy)(2)(x^{2}+y)(2x)}{(x^{2}+y)^{4}}$ = $\frac{(x^{2}+y)(2x^{3}+2y^{2}-6xy-6x^{2}y)}{(x^{2}+y)^{4}}$ = $\frac{2x^{3}+2y^{2}-6xy-6x^{2}y}{(x^{2}+y)^{3}}$
$\frac{\partial w}{\partial y}$ = $\frac{(x^{2}+y)(-1)-(x-y)(1)}{(x^{2}+y)^{2}}$ = $\frac{-x^{2}-x}{(x^{2}+y)^{2}}$
$\frac{\partial^{2} w}{\partial y^{2}}$ = $\frac{-(-x^{2}-x)(2)(x^{2}+y)(1)}{(x^{2}+y)^{4}}$ = $\frac{2x^{2}+2x}{(x^{2}+y)^{3}}$
$\frac{\partial^{2} w}{\partial x{\partial y}}$ = $\frac{(x^{2}+y)^{2}(1+2x)-(-x^{2}+y+2xy)(2)(x^{2}+y)(1)}{(x^{2}+y)^{4}}$ = $\frac{(x^{2}+y)(2x^{3}+2xy+x^{2}+y+2x^{2}-2y-4xy)}{(x^{2}+y)^{4}}$ = $\frac{2x^{3}+3x^{2}-y-2xy}{(x^{2}+y)^{3}}$
