Answer
$\frac{2xdx+2ydy}{(x^2+y^2)^2}$
Work Step by Step
To find the total differential of z, denoted $dz$, use the following formula.
$dz=\frac{dz}{dx}dx+\frac{dz}{dy}dy$
which can also be written as
$dz=f_{x}(x,y)dx+f_{y}(x,y)dy$
Note that these are partial derivatives of f(x,y) in terms of x and y.
$z=\frac{-1}{x^2+y^2}=-(x^2+y^2)^{-1}$
$dz=(x^2+y^2)^{-2}(2x)dx+(x^2+y^2)^{-2}(2y)dy$
$=\frac{2xdx+2ydy}{(x^2+y^2)^2}$
