Reaching upon $9=1$ while solving $x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$

You got $$0\cdot\tan^2{y}+9=1.$$ Id est, you got that $\tan{y}$ does not exist.

Thus, $$y=\frac{\pi}{2}+\pi n,$$ where $n\in\mathbb{Z},$ or $$x+45^{\circ}=\frac{\pi}{2}+\pi n,$$ which gives $$x=\pi n+\frac{\pi}{4}.$$ We did not get a contradiction in Math!

Hint:

Set $\tan y=\dfrac1a$

$$\dfrac{3(1-\sqrt3a)}{a+\sqrt3}=\dfrac{\sqrt3-a}{\sqrt3a+1}$$

$$\iff3-a^2=3(1-3a^2)\iff a=0$$

Another way to avoid confusion:

$$\dfrac31=\dfrac{\sin(x+15^\circ)\cos(x-15^\circ)}{\cos(x+15^\circ)\sin(x-15^\circ)}$$

Apply Componendo et Dividendo

$$\dfrac{3+1}{3-1}=\dfrac{\sin2x}{\sin30^\circ}$$