Answer
inside: $g(x)=16 x^{2}+37 x$
outside: $f(g)=\ln g$
derivative: $f^{\prime}(x)=\dfrac{32x+37 }{16 x^{2}+37 x} $
Work Step by Step
Given $$f(x)=\ln \left(16 x^{2}+37 x\right)$$
Use the chain rule to take the derivative
$$
\frac{d f(g(x))}{d x}=f^{\prime}(g(x)) g^{\prime}(x)
$$
Here $g(x)=16 x^{2}+37 x$ and $f(g)=\ln g,$ then
\begin{align*}
f^{\prime}(x) &=\left(\ln g\right)^{\prime} \\
&=\frac{g^{\prime}(x)}{g (x)} \\
&=\frac{32x+37 }{16 x^{2}+37 x}
\end{align*}