\(\bf \begin{cases}
f(x)=6x\\
g(x)=4x+1
\end{cases}\qquad (f\circ g)(x)\iff f(\quad g(x)\quad )
\\\\\\
\textit{that is}\implies
\begin{cases}
f(\boxed{x})=6\boxed{x}
\\\\
f(\boxed{g(x)})=6\boxed{g(x)}\\
--------------\\
g(x)=4x+1\qquad thus\\
--------------\\
f(\boxed{g(x)})=6(\boxed{4x+1})\iff (f\circ g)(x)
\end{cases}\)
distribute and simplify if needed, see what you get
