Proof by induction on Fibonacci numbers: show that $f_n\mid f_{2n}$

From the start, there isn't a clear statement to induct on. As such, you have to guess the induction hypothesis, and find an explicit pattern which you could describe.

Hint: Look at the sequence of values of $\frac{f_{2k}}{f_k}$. Do you see a pattern there? That suggests to prove the following fact:

$$ \frac{f_{2k+2}} { f_{k+1} } = \frac{f_{2k} } { f_k } + \frac{f_{2k-2} } {f_{k-1} } $$

Check that the first two terms of this series $g_n = \frac{f_{2n}}{f_n}$ are integers, hence conclude by induction that every term is an integer.

The question is old, Calvin Lin's answer is great and already accepted but here is another method (for the famous sake of completess):

We know that $f_n \wedge f_m=f_{n\wedge m}$, where $a\wedge b$ is the gcd of $a$ and $b$. So $f_n \wedge f_{2n}=f_{n\ \wedge\ 2n}=f_n$. This means that $f_n$ divides $f_{2n}$.