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:
$$\frac{3-x}{(x^2+3)(x+3)}=\frac{Ax+B}{x^2+3}+\frac C{x+3} $$ $$= \frac{-\frac12\cdot x+\frac 12}{x^2+3}+\frac {\frac 12}{x+3}=\dfrac 12\left(\frac{1-x}{x^2 + 3}\right) + \dfrac 12\left(\frac 1{x+3}\right)$$
You began by supposing that
$$
\frac{9}{9-x^2}=\frac{a}{3-x}+\frac{b}{3+x}. \tag1
$$
Now you need to find values of $a$ and $b$ that make this true for all $x$ such that $x \neq 3$ and $x \neq -3.$
Next you discover that for every $x$ you care about
(every possible value of $x$ except $3$ and $-3$),
Equation $(1)$ is equivalent to
$$
9= a(3+x) + b(3-x). \tag2
$$
If you can find values of $a$ and $b$ that make Equation $(2)$ true for all $x$ such that $x \neq 3$ and $x \neq -3,$ then you have solved for $a$ and $b$ in Equation $(1)$.
But suppose you can find $a$ and $b$ that Equation $(2)$ true, not just for all $x$ such that $x \neq 3$ and $x \neq -3,$ but for every possible value of $x$
including the possible values $x = 3$ and $x = -3.$
Does the fact that the equation happens to be true for $x=3$ and for $x = -3,$
rather than undefined, give us any a priori reason to doubt that it is true for other values of $x$?
Plugging in $x = 3$ and $x = -3,$ and solving for $a$ and $b$ in each case does not actually (by itself) prove anything about the other values of $x.$
But it does tell us values of $a$ and $b$ that make Equation $(2)$ true for those values of $x$: $a = b = \frac32.$
Now look what happens if you plug these values into the right-hand side of Equation $(2)$:
\begin{align}
a(3+x) + b(3-x) &= \frac32(3+x) + \frac32(3-x) \\
&= \frac32(3) + \frac32 x + \frac32(3) - \frac32 x \\
&= 9
\end{align}
for every possible real number $x,$ including all the possible values that are not $3$ or $-3.$
That's how it works. We really did not care what values of $a$ and $b$ make
$9= a(3+x) + b(3-x)$ when $x = 3$ or when $x = -3.$
Setting $x$ to those values is just a technique for finding out what we need
$a$ and $b$ to be in order to make $9= a(3+x) + b(3-x)$
for all the other possible values of $x.$
And once we have that, we know that
$\frac{9}{9-x^2}=\frac{a}{3-x}+\frac{b}{3+x}$
also will be true for all those values of $x.$
Best Answer
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 $q=s$.