Answer
$$P \equiv(-1,1)$$
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 $P \equiv\left(x_{0}, y_{0}\right)$:
\[
\begin{array}{l}
L(x, y)=-2+2 y-2 x\\
f(x, y)=y^{2}+x^{2}
\end{array}
\]
We know 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 partial derivatives to approximate the function by the Local Approximation Theorem:
\[
\begin{aligned}
f_{x}(x, y) &=\frac{\partial}{\partial x}\left(x^{2}+y^{2}\right) \\
2 x=f_{x}(x, y) &\\
f_{x}\left(x_{0}, y_{0}\right) &=2 x_{0} \\
f_{y}(x, y) &=\frac{\partial}{\partial y}\left(x^{2}+y^{2}\right) \\
2 y=f_{y}(x, y) & \\
f_{y}\left(x_{0}, y_{0}\right) &=2 y_{0}
\end{aligned}
\]
Substituting these values in $L(x, y)$
\[
\begin{array}{l}
2 y-2 x-2=x_{0}^{2}+y_{0}^{2}+2 x_{0}\left(x-x_{0}\right)+2 y_{0}\left(y-y_{0}\right) \\
2 y-2 x-2=-x_{0}^{2}-y_{0}^{2}+2 x_{0} x+2 y_{0} y
\end{array}
\]
Comparing the coefficients of equation (1), we get
\[
\begin{array}{r}
2 x_{0}=-2 \\
2=2 y_{0} \\
-x_{0}^{2}-y_{0}^{2}=-2 \\
x_{0}=-1 \text { and } y_{0}=1
\end{array}
\]
$P \equiv(-1,1)$
