Answer
$\frac{\partial^{2} w}{\partial x^{2}}$ = $-4x^{3}y^{2}[sin(x^{2}y)]+6xy[cos(x^{2}y)]$
$\frac{\partial^{2} w}{\partial y^{2}}$ = $-x^{5}[sin(x^{2}y)]$
$\frac{\partial^{2} w}{\partial x{\partial y}}$ = $-2x^{4}y[sin(x^{2}y)]+3x^{2}[cos(x^{2}y)]$
Work Step by Step
$\frac{\partial w}{\partial x}$ = $(x)cos(x^{2}y)[y(2x)]+sin(x^{2}y)$ = $2x^{2}y[cos(x^{2}y)]+sin(x^{2}y)$
$\frac{\partial^{2} w}{\partial x^{2}}$ = $-2x^{2}y[sin(x^{2}y)](2xy)+[cos(x^{2}y)](4xy)+cos(x^{2}y)[2xy]$ = $-4x^{3}y^{2}[sin(x^{2}y)]+6xy[cos(x^{2}y)]$
$\frac{\partial w}{\partial y}$ = $(x)cos(x^{2}y)[x^{2}]$ = $x^{3}[cos(x^{2}y)]$
$\frac{\partial^{2} w}{\partial y^{2}}$ = $-x^{3}[sin(x^{2}y)](x^{2})$ = $-x^{5}[sin(x^{2}y)]$
$\frac{\partial^{2} w}{\partial x{\partial y}}$ = $-2x^{2}y[sin(x^{2}y)](x^{2})+[cos(x^{2}y)](2x^{2})+cos(x^{2}y)[x^{2}]$ = $-2x^{4}y[sin(x^{2}y)]+3x^{2}[cos(x^{2}y)]$
