Answer
$f_x=\dfrac{-x}{(x^2+y^2+z^2)^{3/2}}$
$f_y=\dfrac{-y}{(x^2+y^2+z^2)^{3/2}}$
$f_z=\dfrac{-z}{(x^2+y^2+z^2)^{3/2}}$
Work Step by Step
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa:
$f_x=-(1/2)(x^2+y^2+z^2)^{-3/2}\times 2x=\dfrac{-x}{(x^2+y^2+z^2)^{3/2}}$
$f_y=-(1/2) \times (x^2+y^2+z^2)^{-3/2}\times 2y=\dfrac{-y}{(x^2+y^2+z^2)^{3/2}}$
$f_z=-(1/2) (x^2+y^2+z^2)^{-3/2}\times 2z=\dfrac{-z}{(x^2+y^2+z^2)^{3/2}}$
