Answer
Prove that $p(1)$ is true.
Prove that $p(k)$ true implies $p(k+1)$ true.
Work Step by Step
Let's note by $p(n)$ the property we want to prove:
$$p(n):\text{ } F_{4n}=3m.$$
The Fibonacci numbers are:
$$1,1,2,3,5,8,13,\dots$$
$\textbf{Step 1:}$ Prove $p(1)$ is true.
Set $n=1$ and check if the property is verified:
$$\begin{align*}
F_4&=F_3+F_2=2+1=3.
\end{align*}$$
As $3$ is divisible by $3$, $p(1)$ is true.
$\textbf{Step 2:}$ Prove that if $p(k)$ is true, then $p(k+1)$ is true.
We assumed $p(k)$ is true:
$$p(k):\text{ } F_{4k}=3m.\tag1$$
We have to prove:
$$p(k+1):\text{ } F_{4(k+1)}=3q.\tag2$$
Rewrite $F_{4(k+1)}$ using the sequence's rule and $F_{4k}=3m$:
$$\begin{align*}
F_{4(k+1)}&=F_{4k+4}\\
&=F_{4k+3}+F_{4k+2}\\
&=F_{4k+2}+F_{4k+1}+F_{4k+1}+F_{4k}\\
&=F_{4k+2}+2F_{4k+1}+F_{4k}\\
&=F_{4k+1}+F_{4k}+2F_{4k+1}+F_{4k}\\
&=3F_{4k+1}+2F_{4k}\\
&=3F_{4k+1}+2(3m)\\
&=3(F_{4k+1}+2m)\\
&=3q.
\end{align*}$$
We got Eq. $(2)$, therefore if $p(k)$ is true, then $p(k+1)$ is also true.
Both steps being proved, the given formula is true for any natural number $n$.
