$\frac{ 2(s^2 + 9(s-1)}{ s (s^2 -9) } = \frac{A}{S} + \frac{ Bs + C}{s^2 -9} $
$ 2(s^2 + 9(s-1) ) = A (s^2 -9) + Bs+C $
Let $S = 0$
$ 2 (-18) = A(-9)$ , so $A = 2$
I sub $A = 2$ into the equation to get
$2s^2 + 18s - 18 = s^2 (2 + B) + Cs - 18$
By comparing coefficients,
$ B = 0$ and $ C= 18$
Why am I wrong here ?
you should factorise the denominator as far as you can while doing partial fraction
$\hspace{10pt}$ $$\frac{2(s^2+9(s-1))}{s(s-3)(s+3)}=\frac{A}{s}+\frac{B}{s-3}+\frac{C}{s+3}$$ $\hspace{10pt}$
multiply both the sides by $s(s-3)(s+3)$ your equations becomes $\hspace{12pt}$
$$2(s^2+9(s-1))=A(s-3)(s+3)+Bs(s+3)+Cs(s-3)$$ $$2s^2+18s-18=A(s^2-9)+B(s^2+3s)+C(s^2-3s)$$ $$2s^2+18s-18=s^2(A+B+C)+s(3B-3C)-9A$$
thus $$A+B+C=2$$ $$3B-3C=18$$ $$-9A=-18 \implies A=2$$
solve for $B, C$ you will get $B=3$,$C=-3$
your partial fraction is $$\frac{2(s^2+9(s-1))}{s(s-3)(s+3)}=\frac{2}{s}+\frac{3}{s-3}-\frac{3}{s+3}$$