1. **State the problem:** Simplify the expression $$\frac{(\frac{1}{2}b + 3a)^2 - (3a + \frac{1}{2}b)(3a - \frac{1}{2}b)}{3 - \frac{1}{3}a(3b + \frac{3}{5}a) + \frac{1}{5}a^2}$$. 2. **Expand the numerator:** - Expand $$\left(\frac{1}{2}b + 3a\right)^2 = \left(\frac{1}{2}b\right)^2 + 2 \cdot \frac{1}{2}b \cdot 3a + (3a)^2 = \frac{1}{4}b^2 + 3ab + 9a^2$$. - Expand $$\left(3a + \frac{1}{2}b\right)\left(3a - \frac{1}{2}b\right) = (3a)^2 - \left(\frac{1}{2}b\right)^2 = 9a^2 - \frac{1}{4}b^2$$ (difference of squares). 3. **Subtract the two expansions in the numerator:** $$\left(\frac{1}{4}b^2 + 3ab + 9a^2\right) - \left(9a^2 - \frac{1}{4}b^2\right) = \frac{1}{4}b^2 + 3ab + 9a^2 - 9a^2 + \frac{1}{4}b^2 = \frac{1}{4}b^2 + \frac{1}{4}b^2 + 3ab = \frac{1}{2}b^2 + 3ab$$. 4. **Simplify the denominator:** - Expand $$- \frac{1}{3}a(3b + \frac{3}{5}a) = - \frac{1}{3}a \cdot 3b - \frac{1}{3}a \cdot \frac{3}{5}a = -ab - \frac{1}{5}a^2$$. - Substitute back: $$3 - ab - \frac{1}{5}a^2 + \frac{1}{5}a^2 = 3 - ab$$ (since $$-\frac{1}{5}a^2 + \frac{1}{5}a^2 = 0$$). 5. **Rewrite the entire expression:** $$\frac{\frac{1}{2}b^2 + 3ab}{3 - ab}$$. 6. **Final answer:** $$\boxed{\frac{\frac{1}{2}b^2 + 3ab}{3 - ab}}$$. This is the simplified form of the given expression.