I managed to find that $\phi(3n) = 3\phi(n) \implies 3|n$ via contradiction:

Suppose that 3 does not divide n. Then gcd(3, n)=1, because 3 is a prime. Which, in turn, implies that $\phi(3n) = \phi(3)\phi(n)$ since Euler's totient function is multiplicative.

Also, note that $\phi(3) = 2$.

However, by hypothesis, $\phi(3n) = 3\phi(n)$, therefore we arrive at the conclusion that $3\phi(n) = 2\phi(n)$

$\phi(n)>0, \forall n \implies 3=2$. Contradiction!

$\therefore 3|n $

However, I can't find my way back. I have tried writing n as $n=3^\alpha m$, where 3 does not divide m. Therefore, $3n = 3^{\alpha+1}m$, which implies that $\phi(3n) = (3^{\alpha+1}-3^\alpha)\phi(m)=3^\alpha(3-1)\phi(m)$ but that only takes me as far as $\phi(3n) = 2.3^\alpha \phi(m)$, and I see no way of progressing (or if that even is the right path).

Does anyone have any suggestions? (Also, it's my first post here, any feedback is very welcome!)