1. **Stating the problem:** We want to find the limit $$\lim_{x \to 0} \frac{\tan 2x - \sin 2x}{3x^3}.$$ 2. **Recall series expansions:** For small $x$, recall that $$\tan y = y + \frac{y^3}{3} + O(y^5)$$ and $$\sin y = y - \frac{y^3}{6} + O(y^5).$$ 3. **Apply expansions for $y=2x$:** \[\tan 2x = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5),\] \[\sin 2x = 2x - \frac{(2x)^3}{6} + O(x^5) = 2x - \frac{8x^3}{6} + O(x^5) = 2x - \frac{4x^3}{3} + O(x^5).\] 4. **Calculate numerator:** \[\tan 2x - \sin 2x = \left(2x + \frac{8x^3}{3}\right) - \left(2x - \frac{4x^3}{3}\right) + O(x^5) = \frac{8x^3}{3} + \frac{4x^3}{3} + O(x^5) = \frac{12x^3}{3} + O(x^5) = 4x^3 + O(x^5).\] 5. **Substitute numerator back into limit:** \[\lim_{x \to 0} \frac{\tan 2x - \sin 2x}{3x^3} = \lim_{x \to 0} \frac{4x^3 + O(x^5)}{3x^3} = \lim_{x \to 0} \frac{4 + O(x^2)}{3} = \frac{4}{3}.\] **Final answer:** $$\boxed{\frac{4}{3}}.$$