Let $g(x)=\sum_{n=0}^\infty a_n x^n$. The main aim here is to get a closed form for the coefficients $b_n$ of $\frac{d \log{g}}{dx}$.
Faà di Bruno's formula tells us that $$\frac{d^n}{dx^n}f(g(x)) = \sum_{k=1}^n f^{(k)}(g(x))B_{n,k}(g'(x),g''(x),...,g^{(n-k+1)})$$
Where $B_{n,k}(g'(x),g''(x),...,g^{(n-k+1)})$ is a partial Bell polynomial.
Moreover, we know that $$B_{n,k}(g'(x),g''(x),...,g^{(n-k+1)})=\frac{n!}{k!} \sum_{\pi_k \in C_n} \prod_{\lambda_j \in \pi_k} \left ( \frac{g^{(\lambda_1)}(x)}{\lambda_1}\right )$$
Where $C_k$ is the set of compositions of $n$ and $\pi_k$ is the composition of $n$ in exactly $k$ parts.
Taking $f=\log$ and $g$ our original power series and applying both Faà di Bruno's formula and Taylor's Theorem, and as we know that $f^{(r)}(g(x))=(-1)^{r-1}(r-1)!g(x)^{-r}$we can get to the conclusion that:
$$\frac{d^n \log{f}}{dx^n} = \sum_{k=1}^n (-1)^{k-1}(k-1)!g(0)^{-k}\frac{n!}{k!} \sum_{\pi_k \in C_n} \prod_{\lambda_j \in \pi_k} \left ( \frac{g^{(\lambda_1)}(x)}{\lambda_1}\right ) $$
Then:
$$b_n=\frac{d^n \log{f}}{dx^n}\frac{1}{n!} = \sum_{k=1}^n \frac{(-1)^{k-1}}{k \ {a_0}^{k}} \sum_{\pi_k \in C_n} \prod_{\lambda_j \in \pi_k} a_{\lambda_j}$$
This can also be seen here and here (section a3, very similar).
The problem comes when I try to work with this formula. As an example, let $g(x)=\sum_{n=0}x^n$. Then,
$$\frac{d \log{g(x)}}{dx} = \frac{\sum_{n=0}nx^{n-1}}{\sum_{n=0}x^n} = \sum_{n=0}x^n$$
So that the coefficients we want are $b_n=1 \ \forall n \in \mathbb{N}$.
On the other hand, applying our formula for, for example, $b_3$, we would get:
$$b_3=\frac{d^3 \log{f}}{dx^3}\frac{1}{3!} = \sum_{k=1}^n \frac{(-1)^{k-1}}{k} \sum_{\pi_k \in C_3} 1= 1+\left ( \frac{-1}{2} \right ) 2 + \frac{1}{3}=\frac{1}{3}$$
Since $C_3=\{\{3\},\{1,2\},\{2,1\},\{1,1,1\}\}$.
Then, what am I doing wrong here? Is there any problem when applying Faà di Bruno's Formula? Am I understanding well the concept of composition? Where is the flaw?
It is no wrong, since $g(x)=\sum^{\infty}_{n=0}x^n=\frac{1}{1-x}$ and if $f(x)=\log(x)$, then $f(g(x))=\log\left(\frac{1}{1-x}\right)=-\log(1-x)=\sum^{\infty}_{n=1}\frac{x^n}{n}$. Hence $b_3=\frac{1}{3}$. Where I used that $\log(1/x)=-\log(x)$, $\forall x>0$.