Answer
$\left(\dfrac{g}{f}\right)(x)=-\dfrac{1}{x+2}$
Domain: $(-\infty,-2)\cup(-2,0.5)\cup(0.5,\infty)$
Work Step by Step
We are given the functions:
$$\begin{align*}
f(x)&=\dfrac{(x+2)^2}{1-2x}\\
g(x)&=\dfrac{x+2}{2x-1}.
\end{align*}$$
Determine $\dfrac{g}{f}$:
$$\begin{align*}
\left(\dfrac{g}{f}\right)(x)&=\dfrac{g(x)}{f(x)}\\
&=\dfrac{\dfrac{x+2}{2x-1}}{\dfrac{(x+2)^2}{1-2x}}\quad&&\text{Substitute the expressions of }f\text{ and }g\\
&=-\dfrac{x+2}{2x-1}\cdot \dfrac{2x-1}{(x+2)^2}\quad&&\text{Invert the divisor and multiply.}\\
&=-\dfrac{1}{x+2}\quad&&\text{Divide numerators and denominators}\\
&\quad&&\text{ by common factors.}
\end{align*}$$
The domain of the function $\dfrac{g}{f}$ is obtained solving the following system of inequalities:
$$\begin{cases}
1-2x&\not=0\\
2x-1&\not=0\\
(x+2)^2&\not=0.
\end{cases}\\
\begin{cases}
x&\not=0.5\\
x&\not=0.5\\
x&\not=-2.
\end{cases}$$
The solution is $(-\infty,-2)\cup(2,0.5)\cup(0.5,\infty)$.
$\left(\dfrac{g}{f}\right)(x)=-\dfrac{1}{x+2}$
Domain: $(-\infty,-2)\cup(-2,0.5)\cup(0.5,\infty)$
