Answer
$w_{xy}=w_{yx}=\dfrac{-6}{(2x+3y)^2}$
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:
$w_x=\dfrac{2}{2x+3y} \\w_y=\dfrac{3}{2x+3y}$
Compute seond partial derivatives:
$w_{xy}=\dfrac{(2x+3y)(0)-(2)(3)}{(2x+3y)^2}=\dfrac{-6}{(2x+3y)^2}$
and
$w_{yx}=\dfrac{(2x+3y)(0)-(3)(2)}{(2x+3y)^2}=\dfrac{-6}{(2x+3y)^2}$
and $w_{xy}=w_{yx}=\dfrac{-6}{(2x+3y)^2}$
