Proving that $n|m\implies f_n|f_m$

Hint: Try to prove that $f_{n+m}=f_{n-1}\ f_m+f_n\ f_{m+1}$.

Then, let $m=n\times k$ for some $k\in \mathbb{N}$.

Now, prove that $f_m$ is divisible $f_n$ by induction on $k$.