The best way to see this: forget about the $X$ in a polynomial or a formal power series, they are really sequences of coefficients, with no constraint on their values, with some specific rules of computation for addition and multiplication.
A power series, however, involves some limiting process, and that requires specific conditions, namely that the series converges.
For instance, you can manipulate $S=\sum_{n=0}^{\infty} x^n$ as a formal power series, and you won't consider convergence, only the operations on it, for instance $S^2=1+2x+3x^2+\dots$. That is, the coefficients of $S$ are $(1,1,1,\dots)$ while the coefficients of $S^2=(1,2,3,\dots)$. But you may also consider $T=\sum_{n=0}^{\infty} n! x^n$, it's a valid formal power series.
Now, for a power series, you require convergence. It's possible to prove that a power series in $x$ converges for all complex number $x$ such that $|x|<R$, for some real (of infinite) $R$. This $R$ is unique and is called the radius of convergence. For instance, the series $S$ above has radius $1$. It converges for $|x|<1$ to the number $\frac{1}{1-x}$. The series $T$ has radius $0$: it never converges if $x\ne0$. As a power series, it's almost useless, but as a formal power series, it can still be useful (we don't care that it does not converge).
There is a similar distinction between a polynom and a polynomial function. But here it's even more tricky, because in the usual undergraduate courses polynom are considered with coefficients in $\Bbb R$ or $\Bbb C$, and there many properties of polynomial directly relate to properties of the associated polynomial function.
When coefficients are in a finite field, it's more surprising. For instance, in $\Bbb F_2$, the finite field with two elements, the polynom $X^2+X$ is not the null polynomial (the null polynomial has null coefficients). However, the function $x\to x^2+x$ only takes the value $0$.
Best Answer
Found the error, turns out I cannot take derivatives of polynomials correctly. Thank you @zwim
Take $$ g(x)=\dfrac{1}{(x-3)^2} $$
and integrate to get $$ \int g(x)dx = \dfrac{-1}{x-3} = \dfrac{1}{3-x} = \dfrac{1}{3}\dfrac{1}{1-\frac{x}{3}} $$
if $u=\frac{x}{3}$ then $$ \dfrac{1}{3}\dfrac{1}{1-u} = \dfrac{1}{3}(1+u+u^2+...) = \dfrac{1}{3}\left(1+\dfrac{x}{3}+\dfrac{x^2}{9}+...\right) $$
Now this is the series for $\int g(x)dx$, so we need to differentiate and get
$$ \dfrac{1}{9}+\dfrac{2x}{27}+\dfrac{3x^2}{81}+... $$
finally we multiply by $4x^2$ to get that
$$ f(x) = \dfrac{1\cdot 4x^2}{9}+\dfrac{2\cdot 4x^3}{27}+... = \sum_{n=0}^\infty \dfrac{(n+1)\cdot 4x^{n+2}}{3^{n+2}} $$