This problem is from Gerstein's Introduction to Mathematical Structures and Proofs: show that if $m|n$ then $\phi(m)|\phi(n)$. One way to approach this is to use the prime factorizations of m and n:
$m=\Pi p_i^{e_i}$
$n=\Pi p_j^{f_j}$
and argue that the $p_j$ include all the $p_i$ with $f_j\ge e_i$
Then
$\phi(m)=\Pi p_i^{e_i-1}\Pi(p_i-1)$
and
$\phi(n)=\Pi p_j^{f_j-1}\Pi(p_j-1)$
You can kind of see that all the factors of $\phi(m)$ are also present in $\phi(n)$, but is there a more direct proof?
The factorizations of $\phi(m)$ and $\phi(n)$ provide a direct proof.
Since if $e_i\leq f_i$, then $e_i-1\leq f_i-1$, and if $p_i$ divides $m$, then $p_i$ divides $n$. This shows that
$\prod_i p^{e_i-1}\prod_i(p_i-1)$ divides $\prod_j p^{e_j-1}\prod_j(p_j-1)$.