Answer
$w_{xx}=4x^2ye^{x^2-y}+2ye^{x^2-y}$
$w_{yy}=y^2e^{x^2-y}-2e^{x^2-y}$
$w_{xy}=w_{yx}=-2xye^{x^2-y}+2xe^{x^2-y}$
Work Step by Step
Our aim is to take the first partial derivative of the given function $w(x,y)$ with respect to $x$, by treating $y$ as a constant and vice versa:
$w_x=2xye^{x^2-y}$
$w_y=-ye^{x^2-y}+e^{x^2-y}$
Now, compute second partial derivatives.
$w_{xx}=4x^2ye^{x^2-y}+2ye^{x^2-y}$
$w_{yy}=y^2e^{x^2-y}-e^{x^2-y}-e^{x^2-y}=y^2e^{x^2-y}-2e^{x^2-y}$
and $w_{xy}=w_{yx}=-2xye^{x^2-y}+2xe^{x^2-y}$
