1. **State the problem:** Multiply $3^{-2}$ by $9^{-3}$. 2. **Recall the properties of exponents:** - $a^{-n} = \frac{1}{a^n}$ - When multiplying powers with the same base, add the exponents: $a^m \times a^n = a^{m+n}$ 3. **Rewrite $9^{-3}$ in terms of base 3:** Since $9 = 3^2$, then $9^{-3} = (3^2)^{-3} = 3^{2 \times (-3)} = 3^{-6}$. 4. **Multiply the expressions:** $$3^{-2} \times 9^{-3} = 3^{-2} \times 3^{-6} = 3^{-2 + (-6)} = 3^{-8}$$ 5. **Simplify the expression:** $$3^{-8} = \frac{1}{3^8}$$ **Final answer:** $$\boxed{\frac{1}{3^8}}$$