= Solution
Since $a_1=f'(0)\ne0$, substitute the proposed inverse series into the forward one. Comparing coefficients of $f,f^2,f^3$ gives
$$
a_1b_1=1,\qquad a_1b_2+a_2b_1^2=0,\qquad
a_1b_3+2a_2b_1b_2+a_3b_1^3=0.
$$
Solving successively yields the <cubic series reversion> coefficients
$$
\boxed{b_1=\frac1{a_1},\qquad b_2=-\frac{a_2}{a_1^3},\qquad
b_3=\frac{2a_2^2-a_1a_3}{a_1^5}.}
$$
The conditions $f(0)=0$ and $a_1\ne0$ ensure that both local series are centred at zero and that the inverse has a linear leading term.
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