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?

Suppose we have
$$f(z)^3 + 2f(z)^2 + f(z) = z$$
and we seek $[z^n] f(z).$ 
While we wait  for a contribution from LIF experts  in the meantime we
can use  Poor Man's Lagrange  Inversion which is the  Cauchy Residue
Theorem. We have
$$[z^n] f(z)
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} f(z) \; dz.$$
Now put $w=f(z)$ so that $w^3+2w^2+w = z$ and
$3w^2 + 4w + 1 \; dw = dz$ to get
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n+1} (1+w)^{2n+2}}
\times w \times (3w^2 + 4w + 1) \; dw.$$
Now $$w \times (3w^2 + 4w + 1) 
= 3 (1+w)^3 - 5 (1+w)^2 + 2 (w+1).$$
Extracting coefficients we thus have
$$3 (-1)^n {n+2n-2\choose n}
- 5 (-1)^n {n+2n-1\choose n}
+ 2 (-1)^n {n+2n\choose n}
\\ = (-1)^n \times
\left(3{3n-2\choose n}
- 5 {3n-1\choose n}
+ 2{3n\choose n}\right).$$
This is
$$(-1)^n {3n-2\choose n}
\left(3 - 5\frac{3n-1}{2n-1}
+ 2 \frac{(3n-1)3n}{(2n-1)2n}\right)
\\ = \frac{(-1)^n}{1-2n} {3n-2\choose n}.$$
Remark.  The fact  that we  are given  $f(0) =  0$  determines the
choice of branch  so that $z=0$ corresponds to  $w=0$ (which is needed
when  we  make  the  substitution).   Also, if  we  choose  $\epsilon$
sufficiently  small  we  also get  a  small  circle  for $w$  (with  a
different $\epsilon.$)
Addendum, five years later. A better choice of integral
is
$$[z^n] f(z) = \frac{1}{n} [z^{n-1}] f'(z)
= \frac{1}{n} \frac{1}{2\pi i}
\int_{|z|=\varepsilon}
\frac{1}{z^n} f'(z) \; dz.$$
We put $w = f(z)$ and obtain
$$\frac{1}{n} \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^n (1+w)^{2n}} \; dw
= \frac{(-1)^{n-1}}{n} {n-1+2n-1\choose n-1}
\\ = \frac{(-1)^{n-1}}{n} {3n-2\choose n-1}.$$
