16 = $4^2$ = $(2^2)^2 = 2^{2\cdot 2} = 2^4$ , $81 = 9^2 = 3^4$. Quindi il tutto diventa
$ \left [ \dfrac{2^{12}}{3^{12}} \div (-1)^2 \dfrac{2^4}{3^4} \right ]^2 \div \left (\dfrac{2}{3} \right )^{14}$
$\left [ \left (\dfrac{2}{3}\right )^{12} \div \left (\dfrac{2}{3} \right )^4 \right ]^2 \div \left (\dfrac{2}{3} \right )^{14}$
$\left [ \left ( \dfrac{2}{3}\right )^{12-4}\right ]^2 \div \left (\dfrac{2}{3}\right )^{14}$
$ \left [ \left ( \dfrac{2}{3}\right )^{8}\right ]^2 \div \left (\dfrac{2}{3} \right )^{14}$
$ \left (\dfrac{2}{3}\right )^{16} \div \left (\dfrac{2}{3}\right )^{14}$
$ \left (\dfrac{2}{3}\right )^{16-14} =\left ( \dfrac{2}{3} \right )^{2} = 4/9 $