We follow here [C.1.1]. Let \(m\) and \(n\) be coprime; we are interested in \(f(mn)\).

Basically, this all boils down to asking what the divisors of \(mn\) look like. Any divisor of \(mn\) must be the product of some divisor \(a\) of \(m\) and some divisor \(b\) of \(n\).

The previous observation is just about multiplication and divisibility, not even coprimeness. But that guarantees that \(a\) and \(b\) are coprime as well, given that \(m\) and \(n\) are. So each divisor \(d\mid mn\) gives us a (unique) pair of (coprime) divisors \(a\) and \(b\) of \(m\) and \(n\).

Instead of summing over all divisors of \(mn\), we can instead sum over each divisor of \(n\) for each divisor of \(m\). In symbols, \begin{equation*}f(mn)=\sum_{a\mid m}\sum_{b\mid n}g(ab)\; .\end{equation*}

Now we can use all the facts we have at hand (coprimeness, multiplicativity, etc.) to finish it off. \begin{equation*}f(mn)=\sum_{a\mid m}\sum_{b\mid n}g(ab)=\sum_{a\mid m}\sum_{b\mid n}g(a)g(b)\end{equation*} \begin{equation*}=\left(\sum_{a\mid m}g(a)\right)\left(\sum_{b\mid n}g(b)\right)=f(m)f(n)\; .\end{equation*}