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

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

Here, \(u = -3x\) and \(n = -\frac{1}{3}\).

First term: \(1\)

Second term: \(-\frac{1}{3}(-3x) = x\)

Third term: \(\frac{-\frac{1}{3}(-\frac{4}{3})}{2}(-3x)^2 = 2x^2\)

Fourth term: \(\frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6}(-3x)^3 = \frac{14}{3}x^3\)

Thus, the expansion up to \(x^3\) is \(1 + x + 2x^2 + \frac{14}{3}x^3\).