Answer
$f(x, y)\approx L(x, y)=3+2x-12y$.
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
$dz=f_{x}(a, b)(x-a)+f_{y}(a, b)(y-b)$
---
$f(x, y)=\displaystyle \frac{2x+3}{4y+1}=(2x+3)(4y+1)^{-1}$
$\begin{array}{ll|ll}
f_{x}(x, y) & =\dfrac{2}{4y+1} & f_{y}(x, y) & =(2x+3)(-1)(4y+1)^{-2}(4)\\
& & & =\dfrac{-8x-12}{(4y+1)^{2}}\\
& & & \\
& & & \\
f_{x}(0,0) & =2 & f_{y}(0,0) & =-12\\
& & &
\end{array}$
$f_{x}$ and $f_{y}$ are continuous functions for $y\displaystyle \neq-\frac{1}{4}$,
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
$dz=f_{x}(a, b)(x-a)+f_{y}(a, b)(y-b)$
$L(x, y)=3+2(x-0)-12(y-0)$
$L(x, y)=3+2x-12y$.
