To solve \(|x - 3| > 2|x + 1|\), consider the critical points where each expression inside the absolute value is zero: \(x = 3\) and \(x = -1\).

Also consider where the expressions \(x - 3\) and \(x + 1\) change sign: \(x = 3\) and \(x = -1\).

Test intervals determined by these points: \((-\infty, -1)\), \((-1, 3)\), and \((3, \infty)\).

1. For \(x < -1\), both \(x - 3\) and \(x + 1\) are negative. The inequality becomes \(-(x - 3) > 2(-(x + 1))\), simplifying to \(x - 3 > 2x + 2\), which gives \(-3 > x + 2\) or \(-5 > x\). This is not possible in this interval.

2. For \(-1 < x < 3\), \(x - 3\) is negative and \(x + 1\) is positive. The inequality becomes \(-(x - 3) > 2(x + 1)\), simplifying to \(x - 3 < -2x - 2\), which gives \(3x < 1\) or \(x < \frac{1}{3}\).

3. For \(x > 3\), both \(x - 3\) and \(x + 1\) are positive. The inequality becomes \(x - 3 > 2(x + 1)\), simplifying to \(x - 3 > 2x + 2\), which gives \(-3 > x + 2\) or \(-5 > x\). This is not possible in this interval.

Combining the valid intervals, the solution is \(-5 < x < \frac{1}{3}\).