Answer
$P \equiv(2,-1)$
Work Step by Step
We are given a function $f(x, y)$ of two variables, and the estimated value of the function by the local approximation theorem at point $P \equiv\left(x_{0}, y_{0}\right)$
\[
\begin{array}{l}
-2+2 y-2 x=L(x, y) \\
x^{2} y=f(x, y)
\end{array}
\]
We know from the local approximation theorem that:
\[
f\left(x_{0}, y_{0}\right)+f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right)=L(x, y)
\]
Therefore, obtaining the partial derivatives
\[
\begin{aligned}
f_{x}(x, y) &=\frac{\partial}{\partial x}\left(x^{2} y\right) \\
2 x y =f_{x}(x, y) &\\
f_{x}\left(x_{0}, y_{0}\right) &=2 x_{0} y_{0} \\
f_{y}(x, y) &=\frac{\partial}{\partial y}\left(x^{2} y\right) \\
f_{y}(x, y) &=x^{2} \\
f_{y}\left(x_{0}, y_{0}\right) &=x_{0}^{2}
\end{aligned}
\]
Subsituting these values in $L(x, y)$ and comparing the coefficients
\[
4 y-4 x+8=x_{0}^{2} y_{0}+2 x_{0} y_{0}\left(x-x_{0}\right)+x_{0}^{2}\left(y-y_{0}\right)
\]
\[
8+4 y-4 x=2 y_{0} x_{0} x+x_{0}^{2} y-x_{0}^{2} y_{0}+x_{0}^{2} y_{0}-2 x_{0}^{2} y_{0}
\]
Thus, we get:
\[
\begin{aligned}
x_{0}^{2} &=4 \\
2 x_{0} y_{0} &=-4 \\
x_{0}^{2} y_{0}-2 x_{0}^{2} y_{0}-x_{0}^{2} y_{0} &=8 \\
x_{0}=2 \text { and } y_{0}=-1 &
\end{aligned}
\]
$P \equiv(2,-1)$
