Answer
\[
P \equiv(1,-1,-1)
\]
Work Step by Step
We are given a function of $f(x, y, z)$ of three variables and the local approximation function $L(x, y)$
\[
\begin{aligned}
&-2+x-y-z =L(x, y, z) \\
x y z=f(x, y, z) &
\end{aligned}
\]
We know that the local approximation function $L(x, y)$ is
\[
f\left(x_{0}, y_{0}\right)+f_{x}\left(x_{0}, y_{0}, z_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}, z_{0}\right)\left(y-y_{0}\right)+f_{z}\left(x_{0}, y_{0}, z_{0}\right)\left(z-z_{0}\right)=L(x, y, z)
\]
We obtain partial derivatives of the function
\[
\begin{aligned}
f_{x}(x, y, z) &=\frac{\partial}{\partial x}(x y z) \\
f_{x}\left(x_{0}, y_{0}, z_{0}\right) &=y_{0} z_{0} \\
f_{y}(x, y, z) &=\frac{\partial}{\partial y}(x y z) \\
f_{y}\left(x_{0}, y_{0}, z_{0}\right) &=x_{0} z_{0} \\
f_{z}(x, y, z) &=\frac{\partial}{\partial z}(x y z) \\
f_{z}\left(x_{0}, y_{0}, z_{0}\right) &=x_{0} y_{0}
\end{aligned}
\]
Substituting these values in $L(x, y, z)$
\[
\begin{array}{rl}
x_{0} y_{0} z_{0}+y_{0} z_{0}\left(x-x_{0}\right)+x_{0} z_{0}\left(y-y_{0}\right)+x_{0} y_{0}\left(z-z_{0}\right)=-2+x-y-z& \\
x\left(y_{0} z_{0}\right)+y\left(x_{0} z_{0}\right)+z\left(x_{0} y_{0}\right)-2 x_{0} y_{0} z_{0}=x-y-z-2 & 2
\end{array}
\]
By comparing coefficients, we get
\[
\begin{aligned}
x_{0} y_{0}=-1, y_{0} z_{0}=1, x_{0} z_{0}=-1 & \text { and } x_{0} y_{0} z_{0}=1 \\
\Rightarrow y_{0}=-1, z_{0}=-1, x_{0}=1
\end{aligned}
\]
$P=(1,-1,-1)$
