It looks like you already have equated coefficients:
$$\frac{3-x}{(x^2+3)(x+3)}=\frac{Ax+B}{x^2+3}+\frac C{x+3}$$
$$\iff \frac{\color{blue}{(Ax+B)(x+3)+C(x^2+3)}}{(x^2+3)(x+3)}= \frac{\color{blue}{3-x}}{(x^2+3)(x+3)}$$
Equating the numerators:
$$(Ax+B)(x+3)+C(x^2+3) = 0\cdot x^2 -x + 3\tag{1}$$
Expanding the LHS of equation $(1)$, gathering like terms:
$$Ax^2 + (B + 3A)x + 3B + Cx^2 + 3C = 0x^2 - x + 3$$
$$\iff \color{blue}{\bf (A + C)}x^2 +\color{red}{\bf (3A + B)}x + \color{green}{\bf (3B + 3C)} = \color{blue}{\bf 0}x^2 + \color{red}{\bf (-1)}x + \color{green}{\bf 3}\tag{2}$$
Match up (color coded above) coefficients on LHS with those on RHS of $(2)$:
$$\iff \color{blue }{\bf A + C = 0}, \quad \color{red}{\bf 3A + B = -1}, \quad \color{green}{\bf 3(B + C) = 3 \iff B+C = 1}$$
Indeed, this gives us a system of $\bf 3$ equations in $\bf 3$ unknowns, from which we can solve for the "unknowns" $A, B, \;\text{and}\; C$.
$$\begin{align} A + C & = 0 \tag{i}\\ 3A + B & = -1 \tag{ii}\\ B+ C & = 1\tag{iii}\end{align}$$
Subtract $(iii)$ from $(i)$: $A - B = -1\tag{iv}$
Adding $(iv)$ to $(ii)$ gives us $4A = -2 \iff A = -\dfrac 12\tag{A}$
From $(i)$: $A = -\dfrac 12 \implies C = \dfrac 12\tag{C}$
From $(iii)$: $C = \dfrac 12 \implies B = \dfrac 12\tag{B}$
Therefore, we have the following function, replacing coefficients A, B, C with their found values:
$$\frac{3-x}{(x^2+3)(x+3)}=\frac{Ax+B}{x^2+3}+\frac C{x+3} $$ $$= \frac{-\frac12\cdot x+\frac 12}{x^2+3}+\frac {\frac 12}{x+3}=\dfrac 12\left(\frac{1-x}{x^2 + 3}\right) + \dfrac 12\left(\frac 1{x+3}\right)$$
Lemma: If $px+q=0$ for all values of $x$, then $p=q=0$.
Proof: In particular, $p(0)+q=0$, which means that $q=0$. So $px=0$ for all $x$, which means that $p(1)=0$, and so $p=0$.
Theorem: If $px+q$ and $rx+s$ are equal for all values of $x$, then $p=r$ and $q=s$.
Proof: If $px+q$ and $rx+s$ are equal for all values of $x$, then
$$
px+q-(rx+s)=0
$$
for all values of $x$. But this expression can be rewritten as
$$
(p-r)x+(q-s)
$$
and so, by the lemma, $p-r=0$ and $q-s=0$. That is, $p=r$ and $q=s$.
Best Answer
HINT:
Observe that the highest power of $x$ are $4,3$ in the numerator & the denominator respectively.
Using Partial Fraction Decomposition,
$$\frac{2x^4-2x^3+x}{(2x-1)^2(x-2)}=Ax+B+\frac C{2x-1}+\frac D{(2x-1)^2}+\frac E{x-2}$$
More generally for
$\displaystyle\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0}=\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0}{\prod (d_ix-c_i)^{n_i}}$
where $m\ge n,$
we can write $\displaystyle\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0}$
$$=e_{m-n}x^{m-n}+e_{m-n-1}x^{m-n-1}+\cdots+e_1x+e_0+\sum\left(\frac {f_{i1}}{d_ix-c_i}+\frac {f_{i2}}{(d_ix-c_i)^2}+\cdots+\frac {f_{i(n_i-1)}}{(d_ix-c_i)^{n_i-1}}+\frac{f_{in_i}}{(d_ix-c_i)^{n_i}}\right)$$