solving a problem with generating functions This is a problem from a course of MIT. 
Find the coefficients of the power series $y = 1 + 3 x + 15 x^2 + 184 x^3 + 495 x^4 + \cdots $ satisfying 
$$ (27 x - 4)y^3 + 3y + 1 = 0 . $$
This is an interesting problem, but I am really cluelss (My naive approaches all failed). Can anyone give me a hint? Or an answer? 

The given coefficients of the power series
\begin{align*}
y(x)=1+3x+15x^2+\color{blue}{84}x^3+495x^4+\cdots 
\end{align*}
indicate they are $\binom{3n}{n}$ resulting in
\begin{align*}
y(x)=\sum_{n=0}^\infty   \binom{3n}{n}x^n \tag{1}
\end{align*}
We prove  (1) by showing  it  fulfils the functional equation
\begin{align*}
(27 x-4)y^3+3y+1=0\tag{2}
\end{align*}
We do so by emplyoing the Lagrange   inversion following the paper  Lagrange Inversion: when and how by  R. Sprugnoli (et al.). We also use  the coefficient of operator  $[x^n]$ to denote the coefficient of $x^n$ of a series. 

We note the coefficients $\binom{3n}{n}$ from (1) are the coefficients of
  \begin{align*}
\binom{3n}{n}=[x^n](1+x)^{3n}\tag{3}
\end{align*}

Let us suppose that a formal power series $w=w(x)$ is implicitely defined by a relation $w=x\Phi(w)$, where $\Phi(x)$ is a formal power series such that $\Phi(0)\ne0$. The Lagrange Inversion Formula (LIF) states that:
$$[x^n]w(x)^k=\frac{k}{n}[x^{n-k}]\Phi(x)^n$$
There are several variations of the LIF stated in the paper. We use in the following formula $G6$ (with $F(x)=1)$:
Let $w=x\Phi(w)$ as before, then the following is valid:
\begin{align*}
[x^n]\Phi(x)^n=\left[\left.\frac{1}{1-x\Phi'(w)}\right|w=x\Phi(w)\right]\tag{4}
\end{align*}
The notation $[\left.f(w)\right|w=g(x)]$ is a linearization of $\left.f(w)\right|_{w=g(x)}$ and denotes the substitution of $g(x)$ to each occurrence of $w$ in $f(w)$ (that is, $f(g(x))$). In particular, $w=x\Phi(w)$ is to be solved in $w=w(x)$ and $w$ has to be substituted in the expression on the left of the $|$ sign.

We    obtain from     (4) with $\Phi(x)=(1+x)^3$ and using
  \begin{align*}
x\Phi^{\prime}(w)=3x(1+w)^2=\frac{3x\Phi(w)}{1+w}=\frac{3w}{1+w}
\end{align*}
\begin{align*}
\binom{3n}{n}&=[x^n](1+x)^{3n}\\
&=[x^n]\left[\left.\frac{1}{1-x\Phi^{\prime}(w)}\right|w=x\Phi(w)\right]\\
&=[x^n]\left[\left.\frac{1}{1-\frac{3w}{1+w}}\right|w=x\Phi(w)\right]\\
&=[x^n]\left[\left.\frac{1+w}{1-2w}\right|w=x\Phi(w)\right]\tag{5}\\
\end{align*} 
It follows from (5)
  \begin{align*}
y(x)=\sum_{n=0}^\infty\binom{3n}{n}x^n=\frac{1+w(x)}{1-2w(x)}
\qquad\text{resp.}\qquad w(x)=\frac{y(x)-1}{2y(x)+1}
\end{align*}
Since $w=x\Phi(w)=x(1+w)^3$, we finally obtain
  \begin{align*}
\frac{y(x)-1}{2y(x)+1)}=x\left(1+\frac{y(x)-1}{2y(x)+1}\right)^3
\end{align*}
  from which 
  \begin{align*}
\color{blue}{(27x-4)y(x)^3+3y(x)+1=0}
\end{align*}
  follows, showing the relationship (2) is fulfilled by (1).

