Answer
$z_{xx}=\frac{16xy+24y^2}{(2x+3y)^4}=\frac{8y}{(2x+3y)^3}$
$z_{yy}=\frac{-24x^2-36xy}{(2x+3y)^4}=\frac{-12x}{(2x+3y)^3}$
$z_{xy}=z_{yx}=\frac{-8x^2+18y^2}{(2x+3y)^4}=\frac{6y-4x}{(2x+3y)^3}$
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:
$z_x=\frac{(2x+3y)(0)-(y)(2)}{(2x+3y)^2}=\frac{-2y}{(2x+3y)^2}$
$z_y=\frac{(2x+3y)(1)-(y)(3)}{(2x+3y)^2}=\frac{2x+3y-3y}{(2x+3y)^2}=\frac{2x}{(2x+3y)^2}$
Then take the derivative of the first order partial derivatives to find second partial derivatives:
$z_{xx}=\frac{(2x+3y)^2(0)-(-2y)(2(2x+3y)\times 2)}{(2x+3y)^4}=\frac{16xy+24y^2}{(2x+3y)^4}$
$z_{yy}=\frac{(2x+3y)^2(0)-(2x)(2(2x+3y)\times 3)}{(2x+3y)^4}=\frac{-24x^2-36xy}{(2x+3y)^4}$
Second partial derivatives of first order partial derivative of x with respect to y and y with respect to x are the same:
$z_{xy}=z_{yx}=\frac{(2x+3y)^2(-2)-(-2y)(2(2x+3y)\times 3)}{(2x+3y)^4}=\frac{-2(4x^2+12xy+9y^2)+24xy+36y^2}{(2x+3y)^4}=\frac{-8x^2-24xy-18y^2+24xy+36y^2}{(2x+3y)^4}=\frac{-8x^2+18y^2}{(2x+3y)^4}$
