1. **State the problem:** Find the first derivative of the function $$y = \ln\left(\frac{3 - x}{(2x + 3)(x + 0)}\right)$$. 2. **Recall the formula:** The derivative of $$\ln(u)$$ with respect to $$x$$ is $$\frac{u'}{u}$$, where $$u$$ is a function of $$x$$. 3. **Define the inner function:** Let $$u = \frac{3 - x}{(2x + 3)(x)}$$. 4. **Simplify the denominator:** $$u = \frac{3 - x}{(2x + 3)x}$$. 5. **Use the quotient rule to find $$u'$$:** $$u' = \frac{(3 - x)' \cdot (2x + 3)x - (3 - x) \cdot ((2x + 3)x)'}{((2x + 3)x)^2}$$ 6. **Calculate derivatives:** $$(3 - x)' = -1$$ $$((2x + 3)x)' = (2x + 3)' \cdot x + (2x + 3) \cdot x' = 2x + 3 + 2x = 4x + 3$$ 7. **Substitute back:** $$u' = \frac{-1 \cdot (2x + 3)x - (3 - x)(4x + 3)}{((2x + 3)x)^2}$$ 8. **Expand numerator:** $$- (2x + 3)x - (3 - x)(4x + 3) = - (2x^2 + 3x) - (12x + 9 - 4x^2 - 3x)$$ $$= -2x^2 - 3x - 12x - 9 + 4x^2 + 3x = ( -2x^2 + 4x^2 ) + ( -3x - 12x + 3x ) - 9 = 2x^2 - 12x - 9$$ 9. **Write $$u'$$:** $$u' = \frac{2x^2 - 12x - 9}{((2x + 3)x)^2}$$ 10. **Write the derivative of $$y$$:** $$y' = \frac{u'}{u} = \frac{\frac{2x^2 - 12x - 9}{((2x + 3)x)^2}}{\frac{3 - x}{(2x + 3)x}} = \frac{2x^2 - 12x - 9}{((2x + 3)x)^2} \cdot \frac{(2x + 3)x}{3 - x}$$ 11. **Simplify:** $$y' = \frac{2x^2 - 12x - 9}{(2x + 3)x} \cdot \frac{1}{3 - x} = \frac{2x^2 - 12x - 9}{(2x + 3)x(3 - x)}$$ **Final answer:** $$\boxed{y' = \frac{2x^2 - 12x - 9}{(2x + 3)x(3 - x)}}$$