Answer
$$y = \frac{1}{{1 - k\left( {\csc x - \cot x} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} - \frac{{{y^2} - y}}{{\sin x}} = 0 \cr
& {\text{separating the variables}} \cr
& \frac{{dy}}{{dx}} = \frac{{{y^2} - y}}{{\sin x}} \cr
& \frac{{dy}}{{{y^2} - y}} = \frac{{dx}}{{\sin x}} \cr
& \frac{{dy}}{{y\left( {y - 1} \right)}} = \csc xdx \cr
& \int {\frac{{dy}}{{y\left( {y - 1} \right)}}} = \int {\csc x} dx \cr
& \int {\frac{{dy}}{{y\left( { - 1 + y} \right)}}} = \int {\csc x} dx \cr
& \cr
& {\text{Integrate by using the formulas }}\int {\frac{{du}}{{u\left( {a + bu} \right)}} = \frac{1}{a}\ln \left| {\frac{u}{{a + bu}}} \right| + C} \cr
& \,\,\,\,\,\,\,\,\,\int {\csc x} dx = \ln \left| {\csc x - \cot x} \right| + C \cr
& {\text{then}}{\text{,}} \cr
& \frac{1}{{ - 1}}\ln \left| {\frac{y}{{ - 1 + y}}} \right| = \ln \left| {\csc x - \cot x} \right| + C \cr
& - \ln \left| {\frac{y}{{y - 1}}} \right| = \ln \left| {\csc x - \cot x} \right| + C \cr
& \ln \left| {\frac{{y - 1}}{y}} \right| = \ln \left| {\csc x - \cot x} \right| + C \cr
& \cr
& {\text{Solving the equation for }}y \cr
& {e^{\ln \left| {\frac{{y - 1}}{y}} \right|}} = {e^{\ln \left| {\csc x - \cot x} \right|}}{e^C} \cr
& \frac{{y - 1}}{y} = k\left( {\csc x - \cot x} \right) \cr
& y - 1 = ky\left( {\csc x - \cot x} \right) \cr
& y - ky\left( {\csc x - \cot x} \right) = 1 \cr
& y = \frac{1}{{1 - k\left( {\csc x - \cot x} \right)}},\,\,\,{\text{where }}k{\text{ is a constant}} \cr} $$