Answer
See below
Work Step by Step
$$\frac{\frac{3-2x}{x^3}}{\frac{2}{x^2}-\frac{1}{x^3+x^2}}\\ =\frac{\frac{3-2x}{x^3}}{\frac{2}{x^2}-\frac{1}{x^2(x+1)}}$$
Multiply all terms by $x^3(x+1)$
$$\frac{\frac{3-2x}{x^3}}{\frac{2}{x^2}-\frac{1}{x^2(x+1)}}\\=\frac{\frac{3-2x}{x^3}}{\frac{2}{x^2}-\frac{1}{x^2(x+1)}}.\frac{x^3(x+1)}{x^3(x+1)}\\=\frac{(3-2x)(x+1)}{2x(x+1)-x}\\=-\frac{(2x-3)(x+1)}{x(2x+1)}$$
