I tried to separate it and found the sum of $$\frac{1}{(1-x/3)^2}$$
but then I got stuck with having to multiply my sum with $(x-1)^2$ . I tried looking online but there's close to nothing about expressing function as power series with a polynomail on the numerator.
I would recommend writing
$$(x-1)^2=(1-x)^2=(3-x-2)^2=(3-x)^2-4(3-x)+4$$
so that
$$\begin{align} {(x-1)^2\over(3-x)^2}&=1-{4\over3-x}+{4\over(3-x)^2}\\ &=1-{4\over3}\cdot{1\over1-x/3}+{4\over9}\cdot{1\over(1-x/3)^2}\\ &=1-{4\over3}\cdot{1\over1-x/3}+{4\over3}\left({1\over1-x/3}\right)' \end{align}$$
If you now plug in the geometric series for $1/(1-x/3)$ and its derivative, you should get something fairly nice.
Another, equivalent, approach:
$$\frac{x-1}{x-3}=1+\frac2{x-3}\implies\left(\frac{x-1}{x-3}\right)^2=\left(1-\frac23\frac1{1-\frac x3}\right)^2=$$
$$=\left(1-\frac23\left(1+\frac x3+\frac{x^2}9+\ldots+\frac{x^n}{3^n}+\ldots\right)\right)^2=\ldots$$
Anyway, the problem of the multiplication of infinite series remains, but as long as you keep it within the convergence radius (in this case, $\;|x|<3\;$) there's no problem
Another way is to expand using partial fractions. The algebra in doing this is easier if you first make the substution $u = 3-x$:
$$ \frac{(x-1)^2}{(3-x)^2} \;\; =\;\; \frac{(2-u)^2}{u^2} \;\; = \;\; \frac{4 - 4u + u^2}{u^2} \;\; = \;\; \frac{4}{u^2} - \frac{4}{u} + 1$$
$$= \;\; \frac{4}{(3-x)^2} - \frac{4}{3-x} + 1 $$
Now notice that this equals
$$-4\frac{d}{dx} \left(\frac{1}{3-x}\right) \;\; - \;\; 4\left(\frac{1}{3-x}\right) \;\; + \;\; 1 $$
Also, using geometric series ideas we have
$$ \frac{1}{3-x} \;\; = \;\; \frac{1}{3\left(1 - \frac{x}{3}\right)} \;\; = \;\; \frac{\frac{1}{3}}{\left(1 - \frac{x}{3}\right)} $$
$$ = \;\; \frac{1}{3} \left[ 1 + \left(\frac{x}{3}\right) + \left(\frac{x}{3}\right)^2 + \left(\frac{x}{3}\right)^3 + \ldots \right] $$
$$ = \;\; \frac{1}{3} \left[ 1 \; + \; \frac{1}{3}x \; + \; \frac{1}{3^2}x^2 \; + \frac{1}{3^3}x^3 \; + \; \ldots \right] $$
Differentiate term-by-term, plug into what follows Now notice that this equals, and combine like terms (citing absolute convergence if you need justification).
