1. **State the problem:** Simplify the complex fraction $$\frac{\frac{a-b}{a+b} - \frac{b}{a-b}}{1 + b \left(-\frac{2}{a+b} - \frac{3}{b-a}\right)}$$ 2. **Rewrite the denominator inside the parentheses:** Note that $b - a = -(a - b)$, so $$-\frac{3}{b-a} = -\frac{3}{-(a-b)} = \frac{3}{a-b}$$ Thus the denominator becomes $$1 + b \left(-\frac{2}{a+b} + \frac{3}{a-b}\right)$$ 3. **Simplify the numerator:** Find a common denominator for the numerator terms $\frac{a-b}{a+b}$ and $\frac{b}{a-b}$, which is $(a+b)(a-b)$: $$\frac{(a-b)^2}{(a+b)(a-b)} - \frac{b(a+b)}{(a-b)(a+b)} = \frac{(a-b)^2 - b(a+b)}{(a+b)(a-b)}$$ 4. **Simplify the numerator's numerator:** $$(a-b)^2 - b(a+b) = (a^2 - 2ab + b^2) - (ab + b^2) = a^2 - 2ab + b^2 - ab - b^2 = a^2 - 3ab$$ So numerator is $$\frac{a^2 - 3ab}{(a+b)(a-b)}$$ 5. **Simplify the denominator:** $$1 + b \left(-\frac{2}{a+b} + \frac{3}{a-b}\right) = 1 + b \left(\frac{-2(a-b) + 3(a+b)}{(a+b)(a-b)}\right)$$ Calculate numerator inside parentheses: $$-2(a-b) + 3(a+b) = -2a + 2b + 3a + 3b = ( -2a + 3a ) + ( 2b + 3b ) = a + 5b$$ So denominator is $$1 + b \frac{a + 5b}{(a+b)(a-b)} = \frac{(a+b)(a-b)}{(a+b)(a-b)} + \frac{b(a + 5b)}{(a+b)(a-b)} = \frac{(a+b)(a-b) + b(a + 5b)}{(a+b)(a-b)}$$ 6. **Simplify the denominator's numerator:** $$(a+b)(a-b) + b(a + 5b) = (a^2 - b^2) + (ab + 5b^2) = a^2 - b^2 + ab + 5b^2 = a^2 + ab + 4b^2$$ 7. **Put it all together:** $$\frac{\frac{a^2 - 3ab}{(a+b)(a-b)}}{\frac{a^2 + ab + 4b^2}{(a+b)(a-b)}} = \frac{a^2 - 3ab}{(a+b)(a-b)} \times \frac{(a+b)(a-b)}{a^2 + ab + 4b^2} = \frac{a^2 - 3ab}{a^2 + ab + 4b^2}$$ 8. **Final answer:** $$\boxed{\frac{a^2 - 3ab}{a^2 + ab + 4b^2}}$$ This is the simplified form of the original complex fraction.