1. Initial line angle

A line makes an angle of $30^\circ$ with the positive $x$-axis. So its inclination is $\theta = 30^\circ.$

2. Rotate clockwise by $15^\circ$

Clockwise rotation decreases the angle, so the new inclination is $\theta' = 30^\circ - 15^\circ = 15^\circ.$

3. Slope of the new line

The slope is $m = \tan 15^\circ.$ Using the standard value, $\tan 15^\circ = 2-\sqrt{3}.$

So the new line passes through $(9,0)$ with slope $2-\sqrt{3}$.

4. Equation using point-slope form

Using point $(9,0)$: $y-0=(2-\sqrt{3})(x-9).$ Thus, $y=(2-\sqrt{3})(x-9).$

5. Rearrange to match options

Expand: $y=(2-\sqrt{3})x-9(2-\sqrt{3}).$ Bring terms together: $y-(2-\sqrt{3})x=-9(2-\sqrt{3}).$

Now divide by $-(2-\sqrt{3})$: $x+\frac{y}{\sqrt{3}-2}=9.$

Since $\frac{1}{2-\sqrt{3}}=2+\sqrt{3}, \quad \frac{1}{\sqrt{3}-2}=-(2+\sqrt{3}),$ this form directly matches $\frac{y}{\sqrt{3}-2}+x=9.$

6. Compare with options

This is exactly Option D.

Therefore, the required equation is $\boxed{\frac{y}{\sqrt{3}-2}+x=9}.$