jmc

Writing the left side with 3 as the base, we have ${\left(\frac{1}{9}\right)}^x=(3^{-2}{)}^x=3^{-2x}$, and thus our equation is: $$3^{-2x}=3^{x+3}.$$Then, by setting the exponents equal to each other, we obtain: $$-2x=x+3.$$This gives us $\boxed{x=-1}$