1. **State the problem:** Solve the inequality $$|x-3| > 2|3x+1|.$$\n\n2. **Recall the definition of absolute value:** For any real number $a$, $|a| = a$ if $a \geq 0$, and $|a| = -a$ if $a < 0$.\n\n3. **Rewrite the inequality:** $$|x-3| > 2|3x+1|.$$\n\n4. **Consider cases based on the expressions inside the absolute values:**\n\n- Case 1: $x-3 \geq 0$ and $3x+1 \geq 0$\n- Case 2: $x-3 \geq 0$ and $3x+1 < 0$\n- Case 3: $x-3 < 0$ and $3x+1 \geq 0$\n- Case 4: $x-3 < 0$ and $3x+1 < 0$\n\n5. **Find critical points:**\n\n- $x-3=0 \Rightarrow x=3$\n- $3x+1=0 \Rightarrow x=-\frac{1}{3}$\n\nThese points divide the real line into intervals: $(-\infty, -\frac{1}{3})$, $[-\frac{1}{3}, 3)$, and $[3, \infty)$.\n\n6. **Solve in each interval:**\n\n**Interval 1: $x < -\frac{1}{3}$**\n- $x-3 < 0 \Rightarrow |x-3| = -(x-3) = 3 - x$\n- $3x+1 < 0 \Rightarrow |3x+1| = -(3x+1) = -3x -1$\nInequality becomes:\n$$3 - x > 2(-3x -1)$$\n$$3 - x > -6x - 2$$\nAdd $6x$ to both sides:\n$$3 - x + 6x > -2$$\n$$3 + 5x > -2$$\nSubtract 3 from both sides:\n$$5x > -5$$\nDivide both sides by 5:\n$$x > -1$$\nCheck if this solution is in the interval $x < -\frac{1}{3}$:\n$x > -1$ and $x < -\frac{1}{3}$ means $-1 < x < -\frac{1}{3}$.\n\n**Interval 2: $-\frac{1}{3} \leq x < 3$**\n- $x-3 < 0 \Rightarrow |x-3| = 3 - x$\n- $3x+1 \geq 0 \Rightarrow |3x+1| = 3x + 1$\nInequality becomes:\n$$3 - x > 2(3x + 1)$$\n$$3 - x > 6x + 2$$\nSubtract 2 from both sides:\n$$1 - x > 6x$$\nAdd $x$ to both sides:\n$$1 > 7x$$\nDivide both sides by 7:\n$$x < \frac{1}{7}$$\nCheck if this solution is in the interval $-\frac{1}{3} \leq x < 3$:\n$x < \frac{1}{7}$ and $x \geq -\frac{1}{3}$ means $-\frac{1}{3} \leq x < \frac{1}{7}$.\n\n**Interval 3: $x \geq 3$**\n- $x-3 \geq 0 \Rightarrow |x-3| = x - 3$\n- $3x+1 \geq 0 \Rightarrow |3x+1| = 3x + 1$\nInequality becomes:\n$$x - 3 > 2(3x + 1)$$\n$$x - 3 > 6x + 2$$\nSubtract $6x$ from both sides:\n$$x - 6x - 3 > 2$$\n$$-5x - 3 > 2$$\nAdd 3 to both sides:\n$$-5x > 5$$\nDivide both sides by $-5$ (reverse inequality):\n$$x < -1$$\nCheck if this solution is in the interval $x \geq 3$:\nNo, since $x < -1$ contradicts $x \geq 3$. So no solution here.\n\n7. **Combine solutions:**\n$$-1 < x < -\frac{1}{3} \quad \text{or} \quad -\frac{1}{3} \leq x < \frac{1}{7}.$$\nSince $-\frac{1}{3}$ is included in the second interval, the combined solution is:\n$$-1 < x < \frac{1}{7}.$$\n\n**Final answer:** $$\boxed{-1 < x < \frac{1}{7}}.$$