I have been wrestling with the following problem for some time now. If possible, I would prefer a hint rather than a full solution, because I would like to "solve" it myself.

Let $f(z)$ be a power-series and $[z^n]\{-\}$ denote the $n$'th coefficient. Show that the following holds, whenever $[z^0]{f(z)}=1$:

$$ \exp\Big[\sum_{n,m>0}\sum_{j>0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\} \frac{q^{n+m}}{(nm)}\Big] = \sum_{n\geq 0}[z^n]\{f(z)^{n-1}\}q^n\,. $$

Remark: This identity appears in comparing two generating series of the same geometric invariants computed using different methods.

Edit 1: Alternatively, if one wishes to get rid of the exponential, one can take a logarithm of the equation and derivative with respect to $q$. Then we are left to solve the following:

$$ \Big[\sum_{n,m>0}\sum_{j>0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\} \frac{(n+m)}{(nm)}q^{n+m}\Big]\sum_{l\geq 0}[z^{l}]\{f(z)^{l-1}\}q^l = \sum_{n\geq 0}n[z^{n}]\{f(z)^{n-1}\}q^n\,. $$