Answer
inside: $g(x)=x^{2}+5 x$
outside: $f(g)=g^{1/3}$
derivative: $f^{\prime}(x)=\dfrac{1}{3}(x^{2}+5 x)^{-2/3} (2x+5)$
Work Step by Step
Given $$f(x)=\sqrt[3]{x^{2}+5 x}$$
Rewriting $f(x)$ as
$$f(x)= (x^{2}+5 x)^{1/3}$$
Use the chain rule to take the derivative
$$
\frac{d f(g(x))}{d x}=f^{\prime}(g(x)) g^{\prime}(x)
$$
Here $g(x)=x^{2}+5 x$ and $f(g)=g^{1/3},$ then
\begin{align*}
f'(x) &=\left(g^{1/3}(x)\right)^{\prime}\\
&=\frac{1}{3}g^{-2/3}(x) g^{\prime}(x)\\
&=\frac{1}{3}(x^{2}+5 x)^{-2/3} (2x+5)
\end{align*}
