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