consider a cubic of the form f(x)=$x^3-2x+z$

Is it possible to derive a power series of coefficients for the function $x^y/f(x)$, for some $y=0,1,2...$ that does not require the use of Faà di Bruno's formula, or does not require multiple iterations of some recursive scheme to get the x^n coefficients .

for example:

https://www.wolframalpha.com/input/?i=%281%2F%28x%5E3-2x%2Bz%29%29++about+x%3D0

The 6th coefficient is $((32 - 12 z^2))/z^6$

So how does one derive such functions of z without having to use a cumbersome recursive scheme ?

It is possible to derive such a closed-form expression for the coefficients of $\frac{3x^2-2}{x^3-2x+z}$

$\displaystyle -z^{-m}2^{m+1} + \sum_{n=0} \frac{z^{2-m}z^{2n} (1+m)}{2^{3n-m+2}(n+1)} \binom{3n+1-m}{n} $

for $3n+1-m<0$

so what about other generalizations?

EDIT:

The closest I got for the $\frac{1}{f(x)}$ case, based on symmetric polynomials,

is $(f(z))^2=4z(a^{-2m-1}+b^{-2m-1}+c^{-2m-1})-4(-z)^{-m}((a^{m+2}+b^{m+2}+c^{m+2}))+-2(-z)^{1-m}((a^{m-1}+b^{m-1}+c^{m-1}))+((a^{-2m+2}+b^{-2m+2}+c^{-2m+2}))$

then...

$f(z)/\sqrt{(32-27z^2)}$

is a perfect square , which gives the coefficient

The $a,b,c$ are roots of $f(x)$, which are functions of z and have a finite expressions as a sum of binomials via the Lagrange inversion theorem.

So for $m=3$

the square root of which is the third coefficient of

https://www.wolframalpha.com/input/?i=%281%2F%28x%5E3-2x%2Bz%29%29++about+x%3D0