Identity for the divisor function: $\tau(mn)=\sum\limits_{d\mid(m,n)}\mu(d) \tau(m/d)\tau(n/d)$ Let $\tau$ denote the classical divisor function and $\mu$ be the 
Möbius function.
Then for each pair of integers $n,m$ we have
$$\tau(mn)=\sum_{d\mid(m,n)}\mu(d) \tau(m/d)\tau(n/d),$$ 
where the sum is taken over all positive integer common divisors of $m$ and $n$. I can verify this by using multiplicativity and checking via brute force that it is true when $m,n$ are powers of the same prime.
My question is whether there is a different proof and whether it is part of a bigger family of similar identities.
 A: We seek to show that
$$\tau(mn) = \sum_{d|(m,n)} \mu(d)\tau(m/d)\tau(n/d).$$
Suppose first that there are prime  factors not shared between $m$ and
$n.$  We show  that we  can  in fact  assume  that there  are no  such
factors. Let  $m'|m$ and $n'|n$  be the maximal contribution  from the
shared factors so that for a prime $p|(m,n)$ it does not divide $m/m'$
and $n/n'$.By virtue of multiplicativity we get for the LHS
$$\tau(m')\tau(m/m') \tau(n') \tau(n/n')$$
and for the RHS
$$\sum_{d|(m,n)} \mu(d)
\tau(m'/d) \tau(m/m') \tau(n'/d) \tau(n/n').$$
We   see   that   the   product    of   the   not   shared   component
$\tau(m/m')\tau(n/n')$ cancels.  Therefore we  may assume that $m$ and
$n$ have the same set of prime factors. Let these primes be indexed as
$p_j$, with the exponent of $p_j$ in  $m$ being $v_j$ and in $n$ being
$w_j.$  We have  for the  sum on  the RHS  the product  representation
(these are just the definitions of $\tau$ and $\mu$) 
$$\tau(m)\tau(n)\prod_{j}
\left(1+\mu(p_j)
\frac{\tau(p_j^{v_j-1})\tau(p_j^{w_j-1})}
{\tau(p_j^{v_j})\tau(p_j^{w_j})}\right)
\\ = \tau(m)\tau(n)\prod_{j}
\left(1-\frac{v_j w_j}{(v_j+1)(w_j+1)}\right)
\\ = \tau(m)\tau(n) \prod_{j} \frac{v_j+ w_j+1}{(v_j+1)(w_j+1)}
\\ = \tau(mn)\prod_j \frac{(v_j+1)(w_j+1)}{v_j+w_j+1}
\prod_{j} \frac{v_j+ w_j+1}{(v_j+1)(w_j+1)}
\\ = \tau(mn).$$
This is  the claim. (On the  first line we  have a factor of  $-1$ for
$\mu(d)$ and  we must replace $v_j+1$  and $w_j+1$ in the  products of
$\tau(m)$ and $\tau(n)$  by $v_j$ and $w_j$ to  obtain $\tau(m/d)$ and
$\tau(n/d).$)
