1. Problem statement: Given the equation $$\cot^{-1} y - \tan^{-1} x = \frac{\pi}{6}$$, prove that $$x + y + \sqrt{3}xy = \sqrt{3}$$. 2. Recall the identity: $$\cot^{-1} y = \tan^{-1} \frac{1}{y}$$ for $y \neq 0$. 3. Substitute into the equation: $$\tan^{-1} \frac{1}{y} - \tan^{-1} x = \frac{\pi}{6}$$. 4. Use the formula for the difference of inverse tangents: $$\tan^{-1} a - \tan^{-1} b = \tan^{-1} \left( \frac{a - b}{1 + ab} \right)$$, provided the result is in the correct range. 5. Applying this formula: $$\tan^{-1} \left( \frac{\frac{1}{y} - x}{1 + x \cdot \frac{1}{y}} \right) = \frac{\pi}{6}$$. 6. Simplify the fraction inside the inverse tangent: $$\frac{\frac{1}{y} - x}{1 + \frac{x}{y}} = \frac{\frac{1 - xy}{y}}{\frac{y + x}{y}} = \frac{1 - xy}{y + x}$$. 7. So we have: $$\tan^{-1} \left( \frac{1 - xy}{x + y} \right) = \frac{\pi}{6}$$. 8. Taking tangent on both sides: $$\frac{1 - xy}{x + y} = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$$. 9. Cross-multiplied: $$\sqrt{3} (1 - xy) = x + y$$. 10. Expand: $$\sqrt{3} - \sqrt{3} xy = x + y$$. 11. Rearranged: $$x + y + \sqrt{3} xy = \sqrt{3}$$. 12. Hence proved the required identity.