Answer
$\dfrac{∂f}{∂x}=\dfrac{4}{(x+y+z^2)^2}$
$\dfrac{∂f}{∂y}=\dfrac{4}{(x+y+z^2)^2}$
and $\dfrac{∂f}{∂z}=\dfrac{8z}{(x+y+z^2)^2}$
$\dfrac{∂f}{∂x}|_{(0,-1,1)}=\text{Undefined}$ and $\dfrac{∂f}{∂y}|_{(0,-1,1)}=\text{Undefined}$
and $\dfrac{∂f}{∂z}|_{(0,-1,1)}=\text{Undefined}$
Work Step by Step
Given: $f(x,y,z)=\dfrac{-4}{x+y+z^2}$
We will find the partial derivatives as follows:
$\dfrac{∂f}{∂x}=-4 \times [\dfrac{-1}{(x+y+z^2)^2}]=\dfrac{4}{(x+y+z^2)^2}$
$\dfrac{∂f}{∂y}=-4 \times [\dfrac{-1}{(x+y+z^2)^2}]=\dfrac{4}{(x+y+z^2)^2}$
and $\dfrac{∂f}{∂z}=-4 \times [-\dfrac{2z}{(x+y+z^2)^2}]=\dfrac{8z}{(x+y+z^2)^2}$
$\dfrac{∂f}{∂x}|_{(0,-1,1)}=\dfrac{4}{(0-1+(1)^2)}=\text{Undefined}$ and $\dfrac{∂f}{∂y}|_{(0,-1,1)}=\dfrac{4}{(0-1+(1)^2)}=\text{Undefined}$
and $\dfrac{∂f}{∂z}|_{(0,-1,1)}=\dfrac{8(1)}{(0-1+(1)^2)}=\text{Undefined}$
