As in Timothy Budd's answer let $w=w(q)$ denote the (formal) solution of $q=\frac{w}{f(w)}$.
Let $p$ be another variable and consider the sum \begin{align*} S(p,q):=\sum_{n,m>0} \sum_{j>0} j[z^{n+j}]\{f(z)^n\} [z^{m-j}]\{f(z)^m\} \frac{p^n q^m}{nm} \end{align*}
Write it as \begin{align*} \sum_{n>0} \frac{p^n}{n}\Big(\sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j\Big) \end{align*}
and rewrite the inner sums as
\begin{align*} \sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j &=\frac{1}{w(q)^n}\Big(f(w(q))^n -\sum_{j=0}^{n-1}[z^j]\{f(z)^n\}w(q)^j\Big) -[z^n] \{f(z)^n\}\\ &=\frac{1}{q^n} -\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j} -[z^n] \{f(z)^n\} \end{align*}
Note that \begin{align*} \sum_{n>0} \frac{p^n}{n}[z^n]\{f(z)^n\}=-\log\big(f(w(p))\big) \end{align*} and that \begin{align*} \sum_{n>0} \frac{p^n}{nq^n}=-\log\big(1-\frac{p}{q}\big) \end{align*} The remaining sum can be written as \begin{align*} \sum_{n>0} \frac{p^n}{n}\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j}&=\sum_{j>0}w(q)^{-j}\sum_{n\geq j}\frac{1}{n}[z^{n-j}]\{f(z)^n\}p^n\\ &=\sum_{j>0}w(q)^{-j}\frac{w(p)^j}{j}\\ &=-\log\big(1-\frac{w(p)}{w(q)}\big) \end{align*} where we have used that for $n\geq j$ (by Lagrange-Bürmann) \begin{align*}\frac{1}{n}[z^{n-j}]\{f(z)^n\}=\frac{1}{n}[z^{n-1}]\{z^{j-1}f(z)^n\}=[q^n]\{\frac{w(q)^j}{j}\}\end{align*}
Thus $S(p,q)=-\log\big(1-\frac{p}{q}\big)+\log\big(1-\frac{w(p)}{w(q)}\big)-\log\big(f(w(p))$ and \begin{align*} \exp(S(p,q))=\frac{1}{f(w(p))}\,\frac{1-\frac{w(p)}{w(q)}}{1-\frac{p}{q}}\end{align*} Now \begin{align*} \frac{q-q\frac{w(p)}{w(q)}}{q-p}&=1+\frac{p}{f(w(q)}\frac{f(w(q))-f(w(p)}{q-p}\\ &=1 +\frac{p f^\prime(w(p))w^\prime(p)}{f(w(q))}+O(q-p)\\ &=1+\frac{f(w(p))}{f(w(q))}\frac{p f^\prime(w(p)}{1+pf^\prime(w(p))} +O(q-p) \end{align*}
so that $\exp(S(q,q)$ reduces to $$\exp(S(q,q))=\frac{1}{f(w(q))}\frac{1}{1-qf^\prime(w(q)}\;\;,$$ as desired.