How do I take a fraction to a negative power?

By definition $$a^{-k} = \frac 1{a^k}$$

So $$\left(\frac{2}{\sqrt{6}}\right)^{-2} =\frac 1{\left(\frac{2}{\sqrt{6}}\right)^{2}}=$$

$$\frac 1{\left(\frac {2^2}{\sqrt 6^2}\right)}=\frac {\sqrt 6^2}{2^2}=\frac 64=\frac 32$$

It will help to realize that $(\frac ab)^{-1} = 1/(a/b) = \frac ba$ and that $(\frac ab)^k = \frac {a^k}{b^k}$ to realize that that means $$\left(\frac ab\right)^{-k} = \frac 1{\left(\frac ab\right)^k}= \frac 1{\left(\frac {a^k}{b^k}\right)} = \frac {b^k}{a^k}.$$

(Also $(\frac ab)^{-k} = [(\frac ab)^{-1}]^k = (\frac ba)^k=\frac {b^k}{a^k}$ or that $(\frac ab)^{-k} = \frac {a^{-k}}{b^{-k}} = (1/a^k)/(1/b^k) = \frac {b^k}{a^k}$.)

In any event

$$\left(\frac {2}{\sqrt 6}\right)^{-2} = \left(\frac {\sqrt 6}{ 2}\right)^2 = \frac {\sqrt 6^2}{2^2} = \frac 64 = \frac 32.$$

$$\begin{align} -2\ln \left( \frac{2}{\sqrt 6} \right) &= -2\big( \ln(2)-\ln(\sqrt{6}) \big) \\ &= -2\ln(2)+2\ln(6^{1/2}) \\ &= -2\ln(2)+\ln(2\cdot 3) \\ &=-2\ln(2)+ \big( \ln(2)+\ln(3)\big) \\ &=\ln(3)-\ln(2) \end{align}$$ And for your specific question, remember that $$\left( \frac{a}{b} \right)^{-n}=\left( \frac{b}{a} \right)^n$$

Well you may start by distributing the index since $2$ and $\sqrt 6$ are positive. Thus $$\left(\frac{2}{\sqrt{6}}\right)^{-2}=\frac{2^{-2}}{{\sqrt 6}^{-2}}.$$

Then recall that for any nonzero number $a$ and any negative integer $-n,$ we have $$a^{-n}=\frac {1}{a^n}.$$ Applying this to your expression, we have $$\frac{2^{-2}}{{\sqrt 6}^{-2}}=\frac{\frac {1}{2^2}}{\frac{1}{{\sqrt 6}^2}}=\frac{\frac {1}{4}}{\frac{1}{6}}=\frac{6}{4}=\frac 32.$$