1. **State the problem:** Find the exact value of $\tan\left(-\frac{2\pi}{3}\right)$.\n\n2. **Recall the formula and rules:** The tangent function is defined as $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.\n\n3. **Evaluate sine and cosine at $\theta = -\frac{2\pi}{3}$:**\nSince $\sin(-x) = -\sin(x)$ and $\cos(-x) = \cos(x)$, we have:\n$$\sin\left(-\frac{2\pi}{3}\right) = -\sin\left(\frac{2\pi}{3}\right)$$\n$$\cos\left(-\frac{2\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right)$$\n\n4. **Use known exact values:**\n$$\sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$\n$$\cos\left(\frac{2\pi}{3}\right) = \cos\left(\pi - \frac{\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2}$$\n\n5. **Substitute these values:**\n$$\sin\left(-\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2}$$\n$$\cos\left(-\frac{2\pi}{3}\right) = -\frac{1}{2}$$\n\n6. **Calculate tangent:**\n$$\tan\left(-\frac{2\pi}{3}\right) = \frac{\sin\left(-\frac{2\pi}{3}\right)}{\cos\left(-\frac{2\pi}{3}\right)} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}$$\n\n7. **Simplify the fraction:**\n$$= \frac{\cancel{-}\frac{\sqrt{3}}{2}}{\cancel{-}\frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$$\n\n8. **Divide by a fraction:**\n$$= \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3}$$\n\n**Final answer:**\n$$\boxed{\sqrt{3}}$$