Find coefficient of $x^ny^m$ how can I find the coefficient of $x^ny^m$ of following series 
$$\frac{\log\big(\frac{1}{1-xy}\big)}{(1-x)(1-y)(1-xy)}$$
and 
$$  \frac{\log(\frac{1}{1-x})}{(1-x)(1-y)(1-xy)}$$
where $1 \leq m \leq n$

Use $\log\left(\frac{1}{1-x y}\right) = -\log(1- xy) = - \sum_{n=1}^\infty \frac{1}{n} x^n y^n$, and 
$$
  -\frac{\log(1-x y)}{1-x y} = \sum_{n=1}^\infty x^n y^n \sum_{m=1}^n \frac{1}{n} = \sum_{n=1}^\infty x^n y^n H_n 
$$
Then use
$$
  \frac{ \sum_{n \geqslant 1, m\geqslant 1} c_{n,m} x^n y^m}{(1-x)(1-y)} = 
   \sum_{n=1}^\infty \sum_{m=1}^\infty x^n y^m \sum_{k=1}^n \sum_{\ell=1}^m c_{k,\ell}
$$
In our case $c_{k,\ell} = H_k \delta_{k,\ell}$, and $\sum_{k=1}^n \sum_{\ell=1}^m H_k \delta_{k,\ell} = \sum_{k=1}^{\min(n,m)} H_k$
$$
  -\frac{\log(1-x y)}{1-x y} \cdot \frac{1}{(1-x)(1-y)} = \sum_{n=1}^\infty \sum_{m=1}^\infty x^n y^m \sum_{k=1}^{\min(n,m)} H_k = \sum_{n=1}^\infty \sum_{m=1}^\infty x^n y^m \left( \left(\min(n,m) +1\right)\left( H_{\min(n,m)+1}-1\right) \right)
$$

For the second series, we start with
$$
  [x]^n \frac{-\log(1-x)}{1-x} = H_n \quad [x]^m \frac{1}{1-x y} = y^m 
$$
Therefore
$$
   [x]^n \frac{-\log(1-x)}{1-x} \cdot \frac{1}{1-x y} = \sum_{k=0}^{n-1} H_{n-k} y^k
$$
Then
$$
   [y]^m [x]^n \frac{-\log(1-x)}{1-x} \cdot \frac{1}{1-x y} \cdot \frac{1}{1-y} = \sum_{k=0}^{n-1}H_{n-k} \mathbf{1}\left( k \leqslant m \right) = \sum_{k=0}^{\min(m, n-1)}H_{n-k} 
$$
