To expand \((3 + 2x)^{-3}\), we use the binomial series expansion for negative exponents:

\((1 + u)^{-n} = 1 - nu + \frac{n(n+1)}{2!}u^2 + \cdots\)

Here, \(u = \frac{2}{3}x\) and \(n = 3\).

First term: \(\frac{1}{27}\)

Second term: \(-3 \times \frac{2}{3}x = -2x\)

Third term: \(\frac{3 \times 4}{2} \left(\frac{2}{3}x\right)^2 = \frac{8}{9}x^2\)

Thus, the expansion up to \(x^2\) is:

\(\frac{1}{27} - \frac{2}{27}x + \frac{8}{81}x^2\)