[tex]y=\displaystyle\sum_{n\ge0}a_nx^n[/tex]
[tex]\sin x=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}x^{2n+1}[/tex]
[tex]\implies\sin y=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}\left(\sum_{m\ge0}a_mx^m\right)^{2n+1}[/tex]
Note that with [tex]y(0)=0[/tex], the first sum reduces immediately (after taking out the first term in the sum) to [tex]a_0[/tex], which implies that [tex]a_0=0[/tex].
So we have
[tex]\displaystyle\sum_{n\ge1}na_nx^{n-1}=9\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}\left(\sum_{m\ge1}a_mx^m\right)^{2n+1}+2\sum_{n\ge0}\frac1{n!}x^n[/tex]
[tex]a_1+2a_2x+3a_3x^2+4a_4x^3+\cdots=9\mathbf S+2\left(1+x+\frac12x^2+\frac16x^3+\cdots\right)[/tex]
where we can expand [tex]\mathbf S[/tex] as
[tex]\mathbf S=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}(a_1x+a_2x^2+a_3x^3+\cdots)^{2n+1}[/tex]
[tex]=a_1x+a_2x^2+a_3x^3+\cdots[/tex]
[tex]\,\,-\dfrac1{3!}(a_1x+a_2x^2+a_3x^3+\cdots)^3+\cdots[/tex]
[tex]=a_1x+a_2x^2+\left(a_3-\dfrac{{a_1}^3}{3!}\right)x^3+\cdots[/tex]
Dropping all terms with order greater than 2 (because this is enough to generate 3 nonzero terms) reduces the equation to
[tex]a_1+2a_2x+3a_3x^2=9\left(a_1x+a_2x^2\right)x^3\right)+2\left(1+x+\dfrac12x^2\right)[/tex]
[tex]a_1+2a_2x+3a_3x^2=2+\left(9a_1+2\right)x+\left(9a_2+1\right)x^2[/tex]
[tex]\implies\begin{cases}a_1=2\\2a_2=9a_1+2\\3a_3=9a_2+1\end{cases}\implies a_1=2,a_2=10,a_3=\dfrac{91}3[/tex]
so that the Taylor polynomial approximating the solution [tex]y[/tex] is
[tex]y(x)\approx2x+10x^2+\dfrac{91}3x^3[/tex]
Just to demonstrate that the result is reasonable, I've attached a plot of an approximate solution with higher accuracy (blue) and the one we found (orange) over the interval [tex]-\dfrac\pi6<x<\dfrac\pi6[/tex].
