Answer
$\dfrac{∂f}{∂x}=y+z$
$\dfrac{∂f}{∂y}=x-z$
and $\dfrac{∂f}{∂z}=x-y$
$\dfrac{∂f}{∂x}|_{(0,-1,1)}=0$ and $\dfrac{∂f}{∂y}|_{(0,-1,1)}=-1$
and $\dfrac{∂f}{∂z}|_{(0,-1,1)}=1$
Work Step by Step
Given: $f(x,y,z)=xy+xz-yz$
We will find the partial derivatives as follows:
$\dfrac{∂f}{∂x}=y+z-0=y+z$
$\dfrac{∂f}{∂y}=x+0-z=x-z$
and $\dfrac{∂f}{∂z}=0+x-y=x-y$
$\dfrac{∂f}{∂x}|_{(0,-1,1)}=-1+1=0$ and $\dfrac{∂f}{∂y}|_{(0,-1,1)}=0-1=-1$
and $\dfrac{∂f}{∂z}|_{(0,-1,1)}=0-(-1)=1$
