Consider the series $$\sum_{m,n=1}^{\infty}\frac{f(m,n)}{m^2n^2}$$. In addition, assume that the function $f$ is weakly multiplicative. Then, how could we evaluate the sum. In particular how can we relate it to the Riemann zeta function?
Can we rewrite the series as $\sum_{m,n}\frac{f(m,n)}{n}\sum_n\frac1{n^3}$ Is this justified. On the other hand, can we use infinite products here, since the function $f$ is weakly multiplicative
With $f(m,n) = lcm(m,n)$ then
$$\sum_{m,n} lcm(m,n)m^{-s} n^{-z} = \sum_d \sum_{m,n,gcd(m,n)=1} lcm(dm,dn) ( dm)^{-s} (dn)^{-z}\\ = \sum_d \sum_{m,n,gcd(m,n)=1} dmn (dm)^{-s} (dn)^{-z}=\zeta(s+z-1)\sum_{m,n,gcd(m,n)=1} m^{1-s} n^{1-z}\\ = \frac{\zeta(s+z-1)}{\zeta(s+z-2)}\sum_d \sum_{m,n,gcd(m,n)=1} (dm)^{1-s} (dn)^{1-z}\\= \frac{\zeta(s+z-1)}{\zeta(s+z-2)} \sum_{M,N} M^{1-s} N^{1-z}=\frac{\zeta(s+z-1)}{\zeta(s+z-2)} \zeta(s-1)\zeta(z-1)$$
Thus for $s=z=2$ it diverges as $\frac{\zeta(s+s-1)}{\zeta(s+s-2)} \zeta(s-1)\zeta(s-1)$ has a double pole at $s=2$.
Something similar holds for any $f(m,n)$ multiplicative in both $m,n$ such that $f(md,nd) = g(d) f(m,n)=g(d) f(m,1)f(1,n)$ for $gcd(m,n)=1$