Extract coefficients for a formal power series using Lagrange Inversion Formula Given $f(x)$ is a formal power series that satisfies $f(0) = 0$
$(f(x))^{3} + 2(f(x))^{2} + f(x) - x = 0$
I know that the Lagrange inversion formula states given f(u) & $\varphi(u)$ are formal power series with respect to u, and $\varphi(0) = 1$ then the following is true.
$[x^{n}](f(u(x))) = \frac{1}{n}[u^{n-1}](f'(u)\varphi(u)^{n})$
How do I find the coefficient of $x^{n}$ in $f(x)$ using Lagrange inversion formula?

Here we use a somewhat simpler but equivalent variant of the Lagrange Inversion Formula. (See Theorem A.2 in Analytic Combinatorics by P. Flajolet and R. Sedgewick for the equivalence of the variants).

Lagrange Inversion Formula: 
Let $g(x), f(x)\in x\mathbb{C}[x]$ be inverses: $g(f(x))=x$. If $g(x)=\frac{x}{\phi(x)}$ and 
  $f(x)=x\phi\left(f(x)\right)$, then
  \begin{align*}
[x^n]f(x)=\frac{1}{n}\left[x^{n-1}\right]\left(\phi(x)\right)^n\tag{1}
\end{align*}

The functional relation
\begin{align*}
(f(x))^{3} + 2(f(x))^{2} + f(x) - x = 0
\end{align*}
can be written as $g\left(f(x)\right)=x$ with
\begin{align*}
g(x)&=x^3+2x^2+x\\
&=x(1+x)^2
\end{align*}

We can write $$\phi(x)=\frac{x}{g(x)}=\frac{1}{(1+x)^2}$$ 
and obtain
\begin{align*}
[x^n]f(x)&=\frac{1}{n}\left[x^{n-1}\right]\left(\phi(x)\right)^n\tag{2}\\
&=\frac{1}{n}\left[x^{n-1}\right]\frac{1}{(1+x)^{2n}}\\
&=\frac{1}{n}\binom{-2n}{n-1}\tag{3}\\
&=\frac{(-1)^{n-1}}{n}\binom{3n-2}{n-1}
\end{align*}

Comment:


*

*In (2) we apply the Lagrange Inversion Formula (1)

*In (3) we use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q
\end{align*}
