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:
$$\frac{3-x}{(x^2+3)(x+3)}=\frac{Ax+B}{x^2+3}+\frac C{x+3} $$ $$= \frac{-\frac12\cdot x+\frac 12}{x^2+3}+\frac {\frac 12}{x+3}=\dfrac 12\left(\frac{1-x}{x^2 + 3}\right) + \dfrac 12\left(\frac 1{x+3}\right)$$
Actually, since the degree of the numerator is not smaller than the degree of the denominator, your first step should be to write $x^5$ as$$(x+2)\times\bigl((x-1)^2(x^2-1)\bigr)+4x^3-2x^2-3x+2.$$So$$\frac{x^5}{(x-1)^2(x^2-1)}=x+2+\frac{4x^3-2x^2-3 x+2}{(x-1)^2(x^2-1)}.$$On the other hand,$$x^2-1=(x-1)(x+1)\implies(x-1)^2(x^2-1)=(x-1)^3(x+1)$$and therefore you should try to get $A$, $B$, $C$ and $D$ such that$$\frac{4x^3-2x^2-3 x+2}{(x-1)^2(x^2-1)}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{(x-1)^3}+\frac D{x+1}.$$
Best Answer
A rational function $P(x)/Q(x)$ can be rewritten using Partial Fraction Decomposition:
$$ \frac{P(x)}{Q(x)} = \frac{A_1}{a\,x + b} + \dots + \frac{A_2\,x + B_2}{a\,x^2 + b\,x + c} + \dots $$
where for each factor of $Q(x)$ of the form $(a\,x + b)^m$ introduce terms:
$$ \frac{A_1}{a\,x + b} + \frac{A_2}{(a\,x + b)^2} + \dots + \frac{A_m}{(a\,x + b)^m} $$
and for each factor of $Q(x)$ of the form $\left(a\,x^2 + b\,x + c\right)^m$ introduce terms:
$$ \frac{A_1\,x + B_1}{a\,x^2 + b\,x + c} + \frac{A_2\,x + B_2}{\left(a\,x^2 + b\,x + c\right)^2} + \dots + \frac{A_m\,x + B_m}{\left(a\,x^2 + b\,x + c\right)^m}\,. $$
In light of all this, you have:
$$ \frac{1}{x^3\left(x^2+1\right)} = \frac{A_1}{x} + \frac{A_2}{x^2} + \frac{A_3}{x^3} + \frac{A_4\,x + B_4}{x^2 + 1} $$
i.e.
$$ \frac{1}{x^3\left(x^2+1\right)} = \frac{\left(A_1 + A_4\right)x^4 + \left(A_2 + B_4\right)x^3 + \left(A_1 + A_3\right)x^2 + A_2\,x + A_3}{x^3\left(x^2+1\right)} $$
which turns out to be an identity if and only if:
$$ \begin{cases} A_1 + A_4 = 0 \\ A_2 + B_4 = 0 \\ A_1 + A_3 = 0 \\ A_2 = 0 \\ A_3 = 1 \end{cases} \; \; \; \; \; \; \Leftrightarrow \; \; \; \; \; \; \begin{cases} A_1 = -1 \\ A_2 = 0 \\ A_3 = 1 \\ A_4 = 1 \\ B_4 = 0 \end{cases} $$
from which what you want:
$$ \frac{1}{x^3\left(x^2+1\right)} = -\frac{1}{x} + \frac{1}{x^3} + \frac{x}{x^2+1}\,. $$