1. **Stating the problem:** Evaluate the double integral $$\int_0^1 \int_y^1 xy \cdot 8xy \, dx \, dy$$. 2. **Rewrite the integrand:** The integrand is $$xy \cdot 8xy = 8x^2 y^2$$. 3. **Set up the integral:** $$\int_0^1 \int_y^1 8x^2 y^2 \, dx \, dy$$ 4. **Integrate with respect to $x$ first:** Treat $y$ as a constant. $$\int_y^1 8x^2 y^2 \, dx = 8 y^2 \int_y^1 x^2 \, dx$$ 5. **Compute the inner integral:** $$\int_y^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_y^1 = \frac{1^3}{3} - \frac{y^3}{3} = \frac{1 - y^3}{3}$$ 6. **Substitute back:** $$8 y^2 \cdot \frac{1 - y^3}{3} = \frac{8}{3} y^2 (1 - y^3) = \frac{8}{3} (y^2 - y^5)$$ 7. **Now integrate with respect to $y$ from 0 to 1:** $$\int_0^1 \frac{8}{3} (y^2 - y^5) \, dy = \frac{8}{3} \int_0^1 (y^2 - y^5) \, dy$$ 8. **Compute the integral:** $$\int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3}$$ $$\int_0^1 y^5 \, dy = \left[ \frac{y^6}{6} \right]_0^1 = \frac{1}{6}$$ 9. **Substitute these results:** $$\frac{8}{3} \left( \frac{1}{3} - \frac{1}{6} \right) = \frac{8}{3} \cdot \frac{1}{6} = \frac{8}{18} = \frac{4}{9}$$ **Final answer:** $$\boxed{\frac{4}{9}}$$