The question is to find all polynomials $P$ such that for all $x\in\mathbb{R}$ $$(x-3)P(x+1)-(x+3)P(x-2)=3x(x^2-9).$$
I've tried putting in different values of $x$, namely those which are zeros of the R.H.S., but all I got from that was that $P(1)=P(-3)=0$.
How do I proceed from here?
A partial answer would be
$$P(x)=(x-1)(x+2)Q(x),$$ for some polynomial $Q$ that satisfies $$Q(x)=Q(x-3)+3.$$ First, note that $P(1)=P(-2)=0,$ by evaluating in $x=3,-3.$ So there must exists a polynomial $Q(x)$ such that $$P(x)=(x-1)(x+2)Q(x).$$ Putting this into the equality, we find that $Q$ must satisfy
$$Q(x+1)=Q(x-2)+3.$$ After changing $x+1 $ for $x$ we find
$$Q(x)=Q(x-3)+3.$$ It is easy to verify that $Q(x)=x-a$ is valid for each real $a.$ Not sure if those are the only ones.
$\textbf{EDIT:}$ LeGrandDODOM(see below) already completed this answer where I left it.
Hint: (when you correctly get $P(1)=P(-2)=0$) you know that $P(x)$ has factor $(x-1)(x+2)$ corresponding to the two roots.
So set $P(x)=(x-1)(x+2)Q(x)$ and see what you can find out about $Q$
Since $$\frac{P(x+1)}{x+3}-\frac{P(x-2)}{x-3}=3x,$$ we obtain $$P(x)=(x+2)(x-1)Q(x),$$ where $Q$ is a polynomial and from here $$Q(x+1)-Q(x-2)=3$$
Filling the blank in Magnusseen's answer.
By writing $P(x)=(x-1)(x+2)Q(x)$ it must be that $Q(x)=Q(x-3)+3$.
Taking derivatives on both sides, $Q'(x)=Q'(x-3)$. Either $Q'$ is a non-zero constant or we can consider $z\in \mathbb C$ a root of $Q'$. Then $Q'(z-3)=0$. Using this argument repeatedly, one finds that all the $z-3n$ where $n\in \mathbb N$ are roots of $Q'$, hence $Q'=0$.
To sum up, there is some $c\in \mathbb R$ such that $Q'=c$, hence $Q(x)=cx+d$ for some $d\in \mathbb R$. Plugging this back into $Q(x)=Q(x-3)+3$ and equating coefficients, we must have $c=1$ and $d$ in undetermined.
Conversely, $Q(x)=x+d$ fits the bill, no matter $d$.
