$$\frac{2tg\frac{\pi}{6}}{tg^2\frac{\pi}{6}-1}=-\frac{2tg\frac{\pi}{6}}{1-tg^2\frac{\pi}{6}}=-tg(2\cdot\frac{\pi}{6})=-tg\frac{\pi}{3}=-\sqrt{3}$$.
Ответ:$$-\sqrt{3}$$