To expand \((1 + 3x)^{\frac{2}{3}}\), we use the binomial series expansion for fractional powers:

\((1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \ldots\)

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

First term: \(1\).

Second term: \(\frac{2}{3} \cdot 3x = 2x\).

Third term: \(\frac{\frac{2}{3}(\frac{2}{3}-1)}{2} \cdot (3x)^2 = -x^2\).

Fourth term: \(\frac{\frac{2}{3}(\frac{2}{3}-1)(\frac{2}{3}-2)}{6} \cdot (3x)^3 = \frac{4}{3}x^3\).

Thus, the expansion up to \(x^3\) is \(1 + 2x - x^2 + \frac{4}{3}x^3\).