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).
To prove (i), consider a power series $F=\sum_{n\geq0}a_nx^n$ and suppose it is such that $F'=F$. Since $F'$ is $\sum_{n\geq1}na_nx^{n-1}=\sum_{n\geq0}(n+1)a_{n+1}x^{n}$, that $F$ and $F'$ are equal means exactly that for all $n\geq0$ we have $a_n=(n+1)a_{n+1}$ or, equivalently, $$a_{n+1}=\frac1{n+1}a_n.$$ If we fix $a_0$, there is exactly one sequence $(a_n)$ that verifies this recurrence relation, $$a_n=\frac{1}{n!}a_0, \quad\forall n\geq0.$$
If we set $a_0=1$, we get the solution you want.
Best Answer
It’s good practice working with summations.
Let $F_1(x)=\sum_{n\ge 0}a_nx^n$, $F_2(x)=\sum_{n\ge 0}b_nx^n$, and $F_3(x)=F_1(x)F_2(x)=\sum_{n\ge 0}c_nx^n$, where we know that
$$c_n=\sum_{k=0}^na_kb_{n-k}\;.$$
Now
$$\begin{align*} F_1(x)F_2'(x)&=\sum_{n\ge 0}a_nx^n\sum_{n\ge 0}nb_nx^{n-1}\\ &=\sum_{n\ge 0}a_nx^n\sum_{n\ge 0}(n+1)b_{n+1}x^n\\ &=\sum_{n\ge 0}\sum_{k=0}^na_k(n-k+1)b_{n-k+1}x^n \end{align*}$$
and
$$\begin{align*} F_1'(x)F_2(x)&=\sum_{n\ge 0}na_nx^{n-1}\sum_{n\ge 0}b_nx^n\\ &=\sum_{n\ge 0}(n+1)a_{n+1}x^n\sum_{n\ge 0}b_nx^n\\ &=\sum_{n\ge 0}\sum_{k=0}^n(k+1)a_{k+1}b_{n-k}x^n\;, \end{align*}$$
so the coefficient of $x^n$ in $F_1(x)F_2'(x)+F_1'(x)F_2(x)$ is
$$\begin{align*} &\sum_{k=0}^n(n-k+1)a_kb_{n-k+1}+\sum_{k=0}^n(k+1)a_{k+1}b_{n-k}\\\\ &\quad=(n+1)a_0b_{n+1}+\sum_{k=1}^n(n-k+1)a_kb_{n-k+1}+\sum_{k=0}^{n-1}(k+1)a_{k+1}b_{n-k}+(n+1)a_{n+1}b_0\\\\ &\quad=(n+1)a_0b_{n+1}+\sum_{k=1}^n(n-k+1)a_kb_{n-k+1}+\sum_{k=1}^nka_kb_{n-k+1}+(n+1)a_{n+1}b_0\\\\ &\quad=(n+1)a_0b_{n+1}+\sum_{k=1}^n(n+1)a_kb_{n-k+1}+(n+1)a_{n+1}b_0\\\\ &\quad=(n+1)\sum_{k=0}^{n+1}a_kb_{n+1-k}\\\\ &\quad=(n+1)c_{n+1}\;, \end{align*}$$
which is of course the coefficient of $x^n$ in $$F_3'(x)=\sum_{n\ge 0}(n+1)c_{n+1}x^n\;.$$