We have the equation $\tan^{-1}4x+\tan^{-1}6x=\frac{\pi}{6}$. Using the formula $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$, we get $$\tan^{-1}\left(\frac{4x+6x}{1-(4x)(6x)}\right) = \frac{\pi}{6}$$ $$\tan^{-1}\left(\frac{10x}{1-24x^2}\right) = \frac{\pi}{6}$$

Taking the tangent of both sides, we have $$\frac{10x}{1-24x^2} = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$$

Cross-multiplying, we get $$10x\sqrt{3} = 1-24x^2$$ $$24x^2 + 10\sqrt{3}x - 1 = 0$$ Using the quadratic formula, $$x = \frac{-10\sqrt{3} \pm \sqrt{(10\sqrt{3})^2 - 4(24)(-1)}}{2(24)} = \frac{-10\sqrt{3} \pm \sqrt{300 + 96}}{48} = \frac{-10\sqrt{3} \pm \sqrt{396}}{48}$$ $$x = \frac{-10\sqrt{3} \pm \sqrt{36 \cdot 11}}{48} = \frac{-10\sqrt{3} \pm 6\sqrt{11}}{48} = \frac{-5\sqrt{3} \pm 3\sqrt{11}}{24}$$ So, $x = \frac{-5\sqrt{3} + 3\sqrt{11}}{24}$ or $x = \frac{-5\sqrt{3} - 3\sqrt{11}}{24}$.

We are given the interval $(\frac{-1}{2\sqrt{6}},\frac{1}{2\sqrt{6}})$. Since $\frac{1}{2\sqrt{6}} = \frac{\sqrt{6}}{12} \approx \frac{2.45}{12} \approx 0.204$, we need to check if the solutions lie in this interval. $x_1 = \frac{-5\sqrt{3} + 3\sqrt{11}}{24} \approx \frac{-5(1.732) + 3(3.317)}{24} \approx \frac{-8.66 + 9.951}{24} \approx \frac{1.291}{24} \approx 0.0538$ $x_2 = \frac{-5\sqrt{3} - 3\sqrt{11}}{24} \approx \frac{-5(1.732) - 3(3.317)}{24} \approx \frac{-8.66 - 9.951}{24} \approx \frac{-18.611}{24} \approx -0.775$ Since $x_1 \approx 0.0538$ and $x_2 \approx -0.775$, we have $x_1 \in (\frac{-1}{2\sqrt{6}},\frac{1}{2\sqrt{6}})$ and $x_2 \notin (\frac{-1}{2\sqrt{6}},\frac{1}{2\sqrt{6}})$. Also, we need to check if $1-24x^2 > 0$ for the solution to be valid. For $x_1$, $1-24x_1^2 \approx 1 - 24(0.0538)^2 \approx 1 - 24(0.00289) \approx 1 - 0.069 \approx 0.931 > 0$. For $x_2$, $1-24x_2^2 \approx 1 - 24(-0.775)^2 \approx 1 - 24(0.6006) \approx 1 - 14.4144 \approx -13.4144 < 0$. Therefore, only $x_1$ is a valid solution.

The number of solutions is 1.