Given, $$f(x) = (9-x^2)^{-1/2}$$
I can pull out a ninth,
$$ f(x) = (9-x^2)^{-1/2} = \left(\dfrac{1}{9} \cdot \left(1+\dfrac{-x^2}{9}\right)\right)^{-1/2}=\dfrac{1}{3} \cdot \left(1+\dfrac{-x^2}{9}\right)^{-1/2} $$
Then substitute into the binomial series $\sum^\infty_{k=0}\binom{\alpha}{k}\cdot x^k = (1+x)^\alpha$ like below remembering that I also need to multiply it by a third:
$$\dfrac{1}{3}\cdot\sum^{\infty}_{k=0}\binom{-1/2}{k}\cdot\left(\dfrac{-x^2}{9}\right)^k$$
Then this can be simplified further... not shown.
However, why is something like this not valid?
$$f(x) = (9-x^2)^{-1/2} = (9+-x^2+(1+-1))^{-1/2} = (1+(8+-x^2))^{-1/2}$$
$$\sum^{\infty}_{k=0}\binom{-1/2}{k}\cdot(8+-x^2)^k$$
Nothing, except that the binomial series for $(1+y)^a$ converges when $\lvert y\rvert <1$, so you need $\lvert 8-x^2 \rvert < 1$, rather than $\lvert -x^2/9\rvert<1 $ for (you'll have to check the equality case separately: it may or may not converge, depending on the value of $a$). The set of $x$ where the series converges is therefore different.
(Note in particular that $x=0$ is not included, which may or may not be a problem for you, depending on what you want to do with it.)