The question asks that if $f(n)$ is multiplicative to prove that $f(n)/n\qquad$ is also multiplicative.
This is what I have:
So, $f(n)\quad$ is multiplicative means that if $p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}\qquad$ is the prime-power decomposition of $n$, then
$f(n)=f(p_1^{e_1})f(p_2^{e_2})\cdots f(p_k^{e_k})$
Now I say, let $g(n)=f(n)/n\qquad $ then $g(n)=f(p_1^{e_1})f(p_2^{e_2})\cdots f(p_k^{e_k})/p_1^{e_1}p_2^{e_2}\cdots p_{k}^{e_{k}}\quad$. Which is by definition multiplicative since $gcd(p_1^{e_1},p_2^{e_2},\ldots,p_k^{e_k})=1\qquad$ and $g(p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}) = f(p_1^{e_1}) f(p_2^{e_2}) \cdots f(p_k^{e_k})/p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$.
I can't think of any other thing to do. Thank you in advance!