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