$z(x+y)=(x-y)\\ z=\dfrac{x-y}{x+y}\\ $

$\dfrac{\partial z}{\partial x}=\dfrac{(x+y).1-(x-y).1}{(x+y)^2}\\ =\dfrac{2y}{(x+y)^2}\\ \dfrac{\partial z}{\partial y}=\dfrac{(x+y).-1-(x-y).1}{(x+y)^2}\\ =-\dfrac{2x}{(x+y)^2}\\ $

$(\dfrac{\partial z}{\partial x}-\dfrac{\partial z}{\partial y} )^2\\ =(\dfrac{2y}{(x+y)^2}-\dfrac{-2x}{(x+y)^2})^2 \\= (\dfrac{2y+2x}{(x+y)^2})^2\\ =4(\dfrac{x+y}{(x+y)^2})^2\\=\dfrac{4}{(x+y)^2}\\ $