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
Take the first partial derivatives of the given function. When taking partial derivative with respect to x, treat y as a constant, and vice versa:
$w_x=2xye^{x^2-y}$
$w_y=-ye^{x^2-y}+e^{x^2-y}$
Then take the derivative of the first order partialderivatives to find second partial derivatives:
$w_{xx}=4x^2ye^{x^2-y}+2ye^{x^2-y}$
$w_{yy}=y^2e^{x^2-y}-e^{x^2-y}-e^{x^2-y}=y^2e^{x^2-y}-2e^{x^2-y}$
Second partial derivatives of first order partial derivative of x with respect to y and y with respect to x are the same:
$w_{xy}=w_{yx}=-2xye^{x^2-y}+2xe^{x^2-y}$
