Use, $$(a^m)^n=(a)^{mn}, a^{-m}=\dfrac{1}{a^m}, a^m \times a^n =a^{m+n}$$ and $$a^m \div a^n =(a)^{m-n}$$
Therefore,
$$\left( \left( \dfrac{-2}{3}\right)^{-2}\right)^3\times \left( \dfrac{1}{3}\right)^{-4}\times 3^{-1}\times \dfrac 16=\left( \dfrac{-2}{3}\right)^{(-2)\times 3}\times (3)^4\times \dfrac 13 \times \dfrac 16$$
$$=\left( \dfrac {-2}{3}\right)^6 \times 3^4 \times \dfrac{1}{3}\times \dfrac{1}{2\times 3}$$
$$=\left( \dfrac{3}{-2}\right)^6 \times 3^4 \times \dfrac{1}{3\times 2\times 3}=\dfrac{(3)^6}{(2)^6}\times 3^4 \times \dfrac{1}{2^1 \times 3^2}$$
$$=\dfrac{(3)^{6+4}}{(2)^{6+1}\times 3^2}$$
$$=\dfrac{(3)^{10}}{2^7 \times 3^2}=\dfrac{3^{10-2}}{2^7}=\dfrac{3^8}{2^7}$$