What should my $u$ be for substitution? $$\int \limits_{\frac 16}^{\frac 12} \csc \left( \frac{\pi}{t} \right) \cot \left(\frac{\pi}{t}\right) \, dt$$

Having trouble deciding on a substitution for this integral .
Your integral diverges. On the other hand, $$\int_{1/2}^{1/6} \csc(\pi t)\cot (\pi t) dt$$ converges and has a nice closed form. I'll assume a typo and show this.

Rewriting $\csc$ and $\cot$ in terms of $\sin$ and $\cos$ we obtain $$\int_{1/2}^{1/6}\frac{\cos \pi t}{\sin^2 \pi t}dt=\int_{1/2}^{1/6}\frac{\cos \pi t}{1-\cos^2 \pi t}dt.$$ Now the substitution we need is $u=\cos \pi t \leftrightarrow dt=-1/\pi \displaystyle\frac{du}{\sqrt{1-u^2}},$ which produces $$-\frac1\pi\int_{\sqrt{3}/2}^{0}\frac{u}{(1-u^2)^{3/2}}du=\frac1\pi\left\lvert\frac{1}{\sqrt{1-u^2}}\right\rvert_0^{\sqrt3/2}=\frac2\pi-\frac1\pi=\bbox[5px,border:2px solid #C0A000]{\frac1\pi.} $$
Firstly, you need to write that expression in terms of sine and cosine then you will know how to proceed further from there.