1. **State the problem:** Simplify the expression $$\frac{(-2^{-3})^{-3}}{(2^{-2})^{-4}}$$. 2. **Recall the power of a power rule:** For any base $a$ and exponents $m$ and $n$, $$(a^m)^n = a^{m \times n}$$. 3. **Simplify the numerator:** $$(-2^{-3})^{-3} = (-1 \times 2^{-3})^{-3} = (-1)^{-3} \times (2^{-3})^{-3}$$ Since $(-1)^{-3} = -1$ and $(2^{-3})^{-3} = 2^{-3 \times -3} = 2^9$, we have: $$-1 \times 2^9 = -2^9$$ 4. **Simplify the denominator:** $$(2^{-2})^{-4} = 2^{-2 \times -4} = 2^8$$ 5. **Rewrite the entire expression:** $$\frac{-2^9}{2^8}$$ 6. **Divide powers with the same base:** $$\frac{2^9}{2^8} = 2^{9-8} = 2^1 = 2$$ So the expression becomes: $$-2$$ 7. **Final answer:** $$\boxed{-2}$$