Let: $$ f(x)=\sum_{i=0}^{\infty}a_ix^i\\ g(x)=\sum_{i=0}^{\infty}b_ix^i $$ and $$ \frac{f(x)}{g(x)}=\sum_{i=0}^{\infty}c_ix^i $$ It is given that $a_0=1, a_1=3, a_2=-2, a_3=-10$ and that $f(x)\cdot g(x)=1=1+0x+0x^2+\dots$.
Find $b_0,b_1,b_2,b_3, c_2$.
As far as I know if: $$ f(x)\cdot g(x)=\sum_{i=0}^{\infty}d_ix^i $$ then: $$ d_i=\sum_{k=0}^i a_kb_{i-k} $$ so: $$ 1=\sum_{k=0}^0 a_kb_{i-k}=a_0b_0=b_0\implies b_0=1\\ 0=\sum_{k=0}^1 a_kb_{i-k}=a_0b_1+a_1b_0=b1+3b_0=b_1+3\implies b_1=-3\\ 0=\sum_{k=0}^2 a_kb_{i-k}=a_0b_2+a_1b_1+a_2b_0=b_2-9-2=b_2-11\implies b_2=11\\ 0=\sum_{k=0}^3 a_kb_{i-k}=b_3+a_1b_2+a_2b_1+a_3b_0=b_3+33+6-10=b_3+29\implies b_3=-29 $$
But I'm not sure how to find $c_2$, namely I'm not sure whether: $$ \frac{1}{g(x)}=\frac{1}{\sum_{i=0}^{\infty} b_ix^i} $$ and what is generating functions arithmetic for division.
Use a similar idea $$ f(x)=g(x)\sum_{i=0}^{\infty}c_ix^i\implies a_n=\sum_{k=0}^nb_kc_{n-k} $$ for all $n\geq 0$. In particular $$ \begin{align}a_0&=b_0c_0\tag{0}\\ a_1&=b_0c_1+b_1c_0\tag{1}\\ a_2&=b_0c_2+b_1c_1+b_2c_0\tag{2} \end{align} $$ You know the values of $a_i$ and $b_i$ for $i=0,1,2$, so you can solve the three equations (0), (1) and (2) in order to determine $c_0, c_1$ and $c_2$.