Answer is: option3

\( y - \frac{1}{\sqrt{3}} = \frac{3}{4} \left( x - \frac{\pi}{6} \right) \)

Solution:

The derivative of \( y = \tan x \) is: \[ \frac{dy}{dx} = \sec^2 x \]

Since \( \sec x = \frac{1}{\cos x} \), we first compute \( \cos \frac{\pi}{6} \):

\[ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \] \[ \sec \frac{\pi}{6} = \frac{1}{\cos \frac{\pi}{6}} = \frac{2}{\sqrt{3}} \] \[ \sec^2 \frac{\pi}{6} = \left( \frac{2}{\sqrt{3}} \right)^2 = \frac{4}{3} \]

Thus, the slope of the tangent line at \( x = \frac{\pi}{6} \) is:

\[ m_{\text{tangent}} = \frac{4}{3} \]

The normal line is perpendicular to the tangent, so its slope is the negative reciprocal of \( \frac{4}{3} \):

\[ m_{\text{normal}} = -\frac{3}{4} \] The equation of a line with slope \( m \) passing through \( (x_1, y_1) \) is: \[ y - y_1 = m(x - x_1) \]

Substituting \( x_1 = \frac{\pi}{6} \), \( y_1 = \frac{1}{\sqrt{3}} \), and \( m = -\frac{3}{4} \):

\[ y - \frac{1}{\sqrt{3}} = -\frac{3}{4} \left( x - \frac{\pi}{6} \right) \]

From the given options, the correct answer is:

\[ \boxed{\quad y - \frac{1}{\sqrt{3}} = -\frac{3}{4} \left( x - \frac{\pi}{6} \right)} \]