Answer
\begin{align*}
f'(g(x))&= \frac{-3}{(x+1)^4}\\
g'(f(x))&= \frac{-3x^2}{(x^3+1)^2}
\end{align*}
Work Step by Step
Given $$f(u)=u^{3}, \quad u=g(x)=\frac{1}{x+1}$$
Since
\begin{align*}
f(g(x))&= [g(x)]^3\\
&= \frac{1}{(x+1)^3}\\
g(f(x))&= \frac{1}{f(x)+1}\\
&= \frac{1}{x^3+1}
\end{align*}
Then
\begin{align*}
f'(g(x))&= \frac{-3}{(x+1)^4}\\
g'(f(x))&= \frac{-3x^2}{(x^3+1)^2}
\end{align*}
