Answer
The partial derivatives $f_{x}$ and $f_{y}$ exist near $(3,0)$, and are continuous functions near $(3,0)$, so $f$ is differentiable at $(3,0)$.
(by Theorem 8)
$L(x, y)=\displaystyle \frac{1}{4}x+y+\frac{5}{4}$.
Work Step by Step
$f(x, y)=\sqrt{x+e^{4y}}=(x+e^{4y})^{1/2}$
$ \begin{array}{lllll}
f_{x}(x, y) & =\frac{1}{2}(x+e^{4y})^{-1/2}(1) & ... & f_{y}(x, y) & =\frac{1}{2}(x+e^{4y})^{-1/2}(4e^{4y})\\
& =\frac{1}{2}(x+e^{4y})^{-1/2} & & & =2e^{4y}(x+e^{4y})^{-1/2}\\
& & & & \\
f_{x}(3,0) & =\frac{1}{2}(3+e^{0})^{-1/2} & & f_{y}(3,0) & =2e^{0}(3+e^{0})^{-1/2}\\
& =\frac{1}{4} & & & =1
\end{array}$
The partial derivatives $f_{x}$ and $f_{y}$ exist near $(3,0)$,
and are continuous functions near $(3,0)$, so $f$ is differentiable at $(3,0)$.
(by Theorem 8)
The linearization of $f$ at (a,b):
$f(x, y)\approx L(x, y)=f(a, b)+ dz$, where
$dz=f_{x}(a, b)(x-a)+f_{y}(a, b)(y-b)$
$L(x, y)=f(3,0)+f_{x}(3,0)(x-3)+f_{y}(3,0)(y-0)$
$L(x, y)=2+\displaystyle \frac{1}{4}(x-3)+1(y-0)$
$L(x, y)=2+\displaystyle \frac{1}{4}x-\frac{3}{4}+y$
$L(x, y)=\displaystyle \frac{1}{4}x+y+\frac{5}{4}$.
