Given, $$\left( \dfrac{2}{9}\right)^{3}\times \left( \dfrac{2}{9}\right)^{-6}=\left( \dfrac{2}{9}\right)^{2x-1}$$.
Using law of exponents, $$a^m \times a^n =a^{m+n}$$
Thus, $$\left( \dfrac{2}{9}\right)^{3-6}=\left( \dfrac{2}{9}\right)^{2x-1}$$
$$\Rightarrow \left( \dfrac{2}{9}\right)^{-3}=\left( \dfrac 29\right)^{2x-1}$$
On comparing both the sides we get,
$$\Rightarrow -3=2x-1$$
$$\Rightarrow -2=2x$$
$$\Rightarrow x=-1$$