Answer
The partial derivatives $f_{x}$ and $f_{y}$ exist near ($2,1$), and are continuous functions for $y\neq x$ so $f$ is differentiable at ($2,1$).
(by Theorem 8)
$L(x, y)==\displaystyle \frac{1}{9}x-\frac{2}{9}y+\frac{2}{3}$
Work Step by Step
$f(x, y)=\displaystyle \frac{x}{x+y}$
$\left.\begin{array}{lllll}
f_{x}(x, y) & =\dfrac{1(x+y)-x(1)}{(x+y)^{2}} & ... & f_{y}(x, y) & =x(-1)(x+y)^{-2}\cdot 1\\
& =\dfrac{y}{(x+y)^{2}} & & & =\dfrac{-x}{(x+y)^{2}}\\
& & & & \\
f_{x}(2,1) & =1/9 & & f_{y}(2,1) & =-2/9
\end{array}\right.$
The partial derivatives $f_{x}$ and $f_{y}$ exist near ($2,1$),
and are continuous functions for $y\neq x$ so $f$ is differentiable at ($2,1$).
(by Theorem 8)
The linearization of $f$ at (a,b):
$f(x, y)\approx L(x, y)=f(a, b)+ dz$, where
$dz=f_{x}(a, b)(x-a)+f_{y}(a, b)(y-b)$
$L(x, y)==f(2,1)+f_{x}(2,1)(x-2)+f_{y}(2,1)(y-1)$
$L(x, y)=\displaystyle \frac{2}{3}+\frac{1}{9}(x-2)-\frac{2}{9}(y-1)$
$L(x, y)=\displaystyle \frac{2}{3}+\frac{1}{9}x-\frac{2}{9}-\frac{2}{9}y+\frac{2}{9}$
$L(x, y)==\displaystyle \frac{1}{9}x-\frac{2}{9}y+\frac{2}{3}$