Answer
inside: $g(x)=1+18 e^{0.6 x}$
outside: $f(g)= 12(g(x) )^{-1}$
derivative: $f^{\prime}(x)=-129.6e^{0.6x}(1+18 e^{0.6 x} )^{-2}$
Work Step by Step
Given$$
f(x)=\frac{12}{1+18 e^{0.6 x}}
$$
Rewriting $f(x)$ as $$ f(x)= 12(1+18 e^{0.6 x})^{-1}$$
Use the chain rule to take the derivative
$$
\frac{d f(g(x))}{d x}=f^{\prime}(g(x)) g^{\prime}(x)
$$
Here $g(x)=1+18 e^{0.6 x}$ and $f(g)= 12(g(x) )^{-1},$ then
\begin{align*}
f^{\prime}(x) &=\left(12(g(x) )^{-1}\right)^{\prime} \\
&=-12(g(x) )^{-2}g^{\prime}(x) \\
&=-12(1+18 e^{0.6 x} )^{-2}((0.6)18 e^{0.6 x}) \\
&=-129.6e^{0.6x}(1+18 e^{0.6 x} )^{-2}
\end{align*}