I don't know if this is according to the site rules/practice, but since no one else bites, I will move my comments here in more editable and hopefully also edible form. I will remove my comments now as that seems to be the usual practice.
To prove part i) we start with the observation that the assumption $b_0=0$ implies that the for all positive integers the power series $G(x)^k$ is of the form
$$
G(x)^k=b_1^kx^k+\sum_{n=k+1}^\infty c_{k,n}x^n
$$
for some coefficients $c_{k,n}$. In the language of the notes (the link is given in Bill Dubuque's comment) we have $\deg G(x)^k\ge k$. Therefore the sum
$$
\sum_{n=0}^\infty a_n(G_n(x))^n
$$
converges to a formal power series $H(x)\in\mathbf{C}[[x]]$ with respect to the $I$-adic topology. Here $I$ is the ideal $I=x\mathbf{C}[[x]]$ (see also Prop. 1.1.8 in Dubuque's link). Part i) is now solved.
To do part ii) we need two Lemmas. I don't know, if they have been given in your textbook and/or lecture notes. The first Lemma is easy.
Lemma 1. If $F_1(x)$ and $F_2(x)$ are power series in $\mathbf{C}[[x]]$, and $F_3(x)=F_1(x)F_2(x)$ is their product, then their formal derivatives satisfy the usual 'derivative of the product' formula
$$
F_3'(x)=F_1(x)F_2'(x)+F_1'(x)F_2(x).
$$
If you have problems in proving this result (or finding a proof), please comment, and I will insert one here.
Corollary. If $F(x)\in \mathbf{C}[[x]]$ and $k$ is a positive integer, then
$$
D(F(x)^k)=k F'(x) F(x)^{k-1}.
$$
Proof. This follows from Lemma 1 as usual by induction on $k$.
Lemma 2. If the series
$$
\sum_{n=0}^\infty F_n(x)
$$
converges to a sum $F(x)$ in the ring $\mathbf{C}[[x]]$ w.r.t to the $I$-adic topology (i.e. in the sense of Dubuque's notes), then so does the series
$$
\sum_{n=0}^\infty F_n'(x).
$$
Furthermore, we have the identity
$$
F'(x)=\sum_{n=0}^\infty F_n'(x).
$$
Proof. If $x^\ell$ divides a summand $F_n(x)$, then clearly $x^{\ell-1}$ divides its derivative $F_n'(x)$. In other words, $\deg F_n'(x)=\deg F_n(x)-1$. As we assume that $\deg F_n(x)\to\infty$ as $n\to\infty$, this implies that $\lim_{n\to\infty}\deg F_n'(x)=\infty$, so the series $\sum_{n=0}F_n'(x)$ converges by Prop. 1.1.8. The claim of the Lemma follows from this, because the sequence of coefficients of any power $x^i$ in the sum eventually becomes a constant.
Again, if you want more details here, just ask!
Now we are in a position to finish off part ii). Let $H(x)=F(G(x))$ that we know to exist in the ring $\mathbf{C}[[x]]$ by part i). First
$$
H'(x)=\sum_{n=0}^\infty D(a_n(G(x))^n
$$
by Lemma 2. Here for each $n$ we have $D(a_n(G(x))^n=na_n(G(x))^{n-1}G'(x)$ by our Corollary. Therefore
$$
H'(x)=\sum_{n=1}^\infty na_n(G(x))^{n-1}G'(x).
$$
By applying part i) to the power series $F'(x)$ and $G(x)$ we see that the series on the right hand side is actually $F'(G(x))G'(x)$. This completes the proof of part ii).
Best Answer
I found this solution by modifying episqrt163's excellent answer giving a formal power series proof of $\exp(\log(\frac1{1-x}))=\frac1{1-x}$. \begin{align} \exp(\log(1+x)) &=\sum_{k\ge 0}\frac{1}{k!}\Big(\log(1+x)\Big)^k \\&=\sum_{k\ge 0}\frac1{k!}\left(\sum_{i_1\ge 1}\frac{(-1)^{i_1-1}x^{i_1}}{i_1}\right)\cdots\left(\sum_{i_k\ge 1}\frac{(-1)^{i_k-1}x^{i_k}}{i_k}\right) \\&=\sum_{k\ge 0}\frac1{k!}\sum_{n\ge k}\; x^n\sum_{\substack{i_1+\dots+i_k=n \\ i_1\ge 1,\dots,i_k\ge 1}} \frac{(-1)^{(i_1-1)+\dots+(i_k-1)}}{i_1 i_2\cdots i_k} \\&=\sum_{n\ge 0}x^n\sum_{k=0}^n\frac1{k!}\sum_{\substack{i_1+\dots+i_k=n \\ i_1\ge 1,\dots,i_k\ge 1}}\frac{(-1)^{n-k}}{i_1 i_2\cdots i_k} \end{align} The innermost ranges over compositions of $n$ with exactly $k$ parts. To simplify this unwieldy summation, we will group together all compositions which correspond to the same unordered integer partition.
Let $\lambda=(\lambda_1,\lambda_2,\dots,\lambda_k)$ be an integer partition of $n$ with exactly $k$ parts. Furthermore, define the multiplicity vector $(m_1,\dots,m_n)$ for $\lambda$, where for each $i\in \{1,\dots,n\}$, $m_i$ is the number of parts equal to $i$ in $\lambda$. The number of ordered compositions $(i_1,\dots,i_k)$ corresponding to $\lambda$ when put in sorted order is $$ \frac{k!}{m_1!\cdots m_n!} $$ This implies that we can re-index the summation to range over integer compositions $\lambda$ of $n$ with $k$ parts, as follows: $$ \begin{align} \exp(\log(1+x)) &= \sum_{n\ge 0}x^n\sum_{k=0}^n\frac1{k!}(-1)^{n-k} \sum_{\substack{\lambda \,\vdash n \\ \text{len}(\lambda)=k}} \frac{k!}{m_1!\cdots m_n!}\frac{1}{1^{m_1}\cdots n^{m_n}} \\&= \sum_{n\ge 0}\frac{x^n}{n!}\sum_{k=0}^n (-1)^{n-k} \sum_{\substack{\lambda \,\vdash n \\ \text{len}(\lambda)=k}} \frac{n!}{m_1!1^{m_1}\cdots m_n!n^{m_n}} \end{align} $$ It is well known that $\frac{n!}{m_1!1^{m_1}\cdots m_n!n^{m_n}}$ is equal to the number of permutations with cycle type $\lambda$. Since we are summing over all $\lambda$ with $k$ parts, the innermost sum equals the number of permutations with exactly $k$ cycles, i.e. the unsigned Stirling number of the first kind. That is, $$ \exp(\log(1+x))=\sum_{n\ge 0}\frac{x^n}{n!}\sum_{k=0}^n (-1)^{n-k}{n \brack k} $$ Finally, the alternating sum $\sum_{k=0}^n (-1)^{n-k}{n \brack k}$ is equal to zero for all $n\ge 2$, because it computes the difference between the number of permutations with an even number of cycles and the number of permutations with an odd number of cycles. There is a bijection between these two groups, namely, multiplying by any transposition. The exception is $n=0$ and $n=1$, where $n$ is too small for a transposition to exists. In these cases, the difference is one. This exactly corresponds to the series $$ 1+x+\color{gray}{0x^2+0x^3+\dots}, $$ completing the proof.