Answer
$w_{xy}=w_{yx}=\frac{-6}{(2x+3y)^2}$
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:
$w_x=\frac{1}{2x+3y}\times2=\frac{2}{2x+3y}$
$w_y=\frac{1}{2x+3y}\times3=\frac{3}{2x+3y}$
Then take the derivative of the first order partial derivatives to find second partial derivatives:
$w_{xy}=\frac{(2x+3y)(0)-(2)(3)}{(2x+3y)^2}=\frac{-6}{(2x+3y)^2}$
$w_{yx}=\frac{(2x+3y)(0)-(3)(2)}{(2x+3y)^2}=\frac{-6}{(2x+3y)^2}$
