Answer
$\dfrac{∂f}{∂x}=\dfrac{-12x}{(x^2+y^2+z^2)^{2}}$
$\dfrac{∂f}{∂y}=\dfrac{-12y}{(x^2+y^2+z^2)^{2}}$
and $\dfrac{∂f}{∂z}=\dfrac{-12z}{(x^2+y^2+z^2)^{2}}$
$\dfrac{∂f}{∂x}|_{(0,-1,1)}=0$ and $\dfrac{∂f}{∂y}|_{(0,-1,1)}=3$
and $\dfrac{∂f}{∂z}|_{(0,-1,1)}=-3$
Work Step by Step
Given: $f(x,y,z)=\dfrac{6}{x^2+y^2+z^2}=6(x^2+y^2+z^2)^{-1}$
We will find the partial derivatives as follows:
$\dfrac{∂f}{∂x}=-6(x^2+y^2+z^2)^{-2}\times (2x)=\dfrac{-12x}{(x^2+y^2+z^2)^{2}}$
$\dfrac{∂f}{∂y}=-6(x^2+y^2+z^2)^{-2}\times (2y)=\dfrac{-12y}{(x^2+y^2+z^2)^{2}}$
and $\dfrac{∂f}{∂z}=-6(x^2+y^2+z^2)^{-2}\times (2z)=\dfrac{-12z}{(x^2+y^2+z^2)^{2}}$
$\dfrac{∂f}{∂x}|_{(0,-1,1)}=\dfrac{-12(0)}{[0^2+(-1)^2+(1)^2]^{2}}=0$ and $\dfrac{∂f}{∂y}|_{(0,-1,1)}=\dfrac{-12(-1)}{[0^2+(-1)^2+(1)^2]^{2}}=3$
and $\dfrac{∂f}{∂z}|_{(0,-1,1)}=\dfrac{-12(1)}{[0^2+(-1)^2+(1)^2]^{2}}=-3$
