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