Cannot find $\displaystyle \int_0^{\frac{\pi}{6}} \frac{1}{\sin x-\cos x} \, dx$

HINT:

$$\sin(x)-\cos(x)=\sqrt 2 \sin(x-\pi/4)$$

and integrate the cosecant function.

If you wish to proceed as in the OP, then we have

$$\begin{align} \int_0^{\pi/6}\frac{1}{\sin(x)-\cos(x)}\,dx&=\int_0^{\pi/6}\frac{\sin(x)+\cos(x)}{\sin^2(x)-\cos^2(x)}\,dx\\\\ &=\int_1^{\sqrt 3/2}\frac{1}{2u^2-1}\,du+\int_0^{1/2}\frac{1}{2v^2-1}\,du\\\\ &=\frac12 \int_1^{\sqrt 3/2}\left(\frac{1}{\sqrt 2 u-1}-\frac{1}{\sqrt 2 u+1}\right)\,du+\frac12 \int_0^{1/2}\left(\frac{1}{\sqrt 2 v-1}-\frac{1}{\sqrt 2 v+1}\right)\,dv\\\\\ \end{align}$$

Can you finish now?

If nothing else works, there's always the "universal trig substitution" $t=\tan\frac{x}{2}$, because then, due to trig identities, $\sin x=\frac{2t}{1+t^2}$, $\cos x=\frac{1-t^2}{1+t^2}$, and (just in case) $\tan x=\frac{2t}{1-t^2}$. That will take you to the realm of integrating rational functions, for which there's a standard (albeit lengthy) algorithm.