Answer
$$\frac{\cot \left(x\right)\csc \left(x\right)}{\left(\csc \left(x\right)-1\right)^2}$$
Work Step by Step
Given $$f(u)=\frac{u}{u-1}, \quad g(x)=\csc x$$
Since
\begin{align*}
f(g(x))&=\frac{g(x)}{g(x)-1}\\
&= \frac{\csc x }{\csc x-1}
\end{align*}
Then
\begin{align*}
f'(g(x))&= \frac{\frac{d}{dx}\left(\csc \left(x\right)\right)\left(\csc \left(x\right)-1\right)-\frac{d}{dx}\left(\csc \left(x\right)-1\right)\csc \left(x\right)}{\left(\csc \left(x\right)-1\right)^2}\\
&= \frac{\cot \left(x\right)\csc \left(x\right)}{\left(\csc \left(x\right)-1\right)^2}
\end{align*}
