I'm trying to calculate $$[z^n u^j] \frac{1}{(1-zu)(1-z)} \log \left(\frac{1}{1-zu}\right),$$ where $[z^n u^j] \sum_{n=0}^\infty \sum_{j=0}^\infty F_{n,j} z^n u^j = F_{n,j}$.
So I have to calculate the coefficient of a generating function in two variables for $$\frac{1}{(1-zu)(1-z)} \log\left(\frac{1}{1-zu}\right).$$
As I know the series representation of $\frac{1}{(1-zu)} = \sum_{n=0}^\infty (uz)^n$, $\frac{1}{(1-z)} = \sum_{n=0}^\infty z^n$, $\log \frac{1}{(1-zu)} = \sum_{n=1}^\infty \frac{(uz)^n}{n}$, I tried to build a cauchy product of the first and third series
$$\begin{align} \frac{1}{(1-z)(1-zu)} \log \left(\frac{1}{1-zu}\right)&= \sum_{n=0}^\infty z^n \sum_{n=0}^\infty (uz)^n \sum_{n=1}^\infty \frac{(uz)^n}{n}\\ &= \sum_{n=0}^\infty z^n \sum_{k=0}^\infty \sum_{l=1}^k \frac{1}{l} u^k z^k\\ &= \sum_{n=0}^\infty \sum_{k=0}^\infty H_k u^kz^{k+n} \end{align}$$ but know I don't know how to go on. What could I do to get something in the form of $$\sum_{n=0}^\infty \sum_{j=0}^\infty F_{n,j} z^n u^j\,?$$
You are correct. From your attempt it follows that $$[z^n u^j] \frac{1}{(1-zu)(1-z)} \log \left(\frac{1}{1-zu}\right) =\begin{cases} H_j &\text{if $1\leq j\leq n$}\\ 0 &\text{otherwise} \end{cases} $$