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I would like to prove that for any function $f(x, y)$:

$$\sum\limits_{d\mid n}\sum\limits_{c\mid d}f(c,d) = \sum\limits_{c\mid n}\sum\limits_{d\mid (n/c)}f(c,cd)$$

My thinking is that:

1) I would like to write left hand side in the same form as RHS so:
$$\sum\limits_{cd\mid n}\sum\limits_{c\mid cd}f(c,cd)$$
2) Interchange the summation:
$$\sum\limits_{c\mid cd}\sum\limits_{cd\mid n}f(c,cd)$$
3) Compare lhs and rhs:
$$\sum\limits_{c\mid cd}\sum\limits_{cd\mid n}f(c,cd) = \sum\limits_{c\mid n}\sum\limits_{cd\mid n}f(c,cd)$$

Now I'm stuck $\sum\limits_{c\mid cd}$ is not $\sum\limits_{c\mid n}$ at least for $n

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    That is unusual notation. Please clarify what the summand subscripts actually mean. – 2017-02-08
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    @GrahamKemp $d\mid n$ means "$d$ divides $n$" – 2017-02-08
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    ... in terms of summation intervals. – 2017-02-08
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    @GrahamKemp What intervals? For example, $\sum_{\,d\mid 6}a_d=a_1+a_2+a_3+a_6$ because $1,2,3,6$ are the divisors of $6$. This is standard notation in number theory. – 2017-02-08

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In other words we want an equality

$$ \{(c,d): c\mid d\mid n\}=\{(a,ab): a\mid n,\, b\mid (n/a)\}.$$

Defining $a=c$ and $b=d/c$, this means we must show

$$c\mid d\mid n\iff a\mid n~\wedge~b\mid(n/a).$$

$(\Rightarrow)$: Suppose $c\mid d\mid n$. Then $c\mid n$ so we conclude $a\mid n$. Moreoever, we get $1\mid (d/c)\mid(n/c)$ or more simply just $(d/c)\mid(n/c)$ or in other words $b\mid(n/a)$.

$(\Leftarrow)$: Since $d=bc$ we already know $c\mid d$. Moreoever if $b\mid (n/a)$ then $ab\mid n$, i.e. $d\mid n$. So we conclude that $c\mid d\mid n$.