Answer
$f_w=\frac{z^2-w^2}{(w^2+z^2)^2}$
$f_z=\frac{-2wz}{(w^2+z^2)^2}$
Work Step by Step
Take the first partial derivatives of the given function. When taking partial derivative with respect to w, treat z as a constant, and vice versa:
$f_w=\frac{(w^2+z^2)(1)-(w)(2w+0)}{(w^2+z^2)^2}=\frac{w^2+z^2-2w^2}{(w^2+z^2)^2}=\frac{z^2-w^2}{(w^2+z^2)^2}$
$f_z=\frac{(w^2+z^2)(0)-(w)(0+2z)}{(w^2+z^2)^2}=\frac{-2wz}{(w^2+z^2)^2}$
