Answer
$$y = \frac{{{C_1}{x^3} - 2x}}{2}$$
Work Step by Step
$$\eqalign{
& \left( {2x + 3y} \right)dx - xdy = 0 \cr
& {\text{Let }}y = ux,{\text{ }}dy = udx + xdu \cr
& {\text{Substituting}} \cr
& \left( {2x + 3ux} \right)dx - x\left( {udx + xdu} \right) = 0 \cr
& 2xdx + 3uxdx - uxdx - {x^2}du = 0 \cr
& 2xdx + 2uxdx - {x^2}du = 0 \cr
& {\text{Separate the variables}} \cr
& 2xdx + 2uxdx = {x^2}du \cr
& \left( {2 + 2u} \right)xdx = {x^2}du \cr
& \frac{x}{{{x^2}}}dx = \frac{1}{{2 + 2u}}du \cr
& {\text{Integrate both sides}} \cr
& \ln \left| {Cx} \right| = \frac{1}{2}\ln \left| {2u + 2} \right| \cr
& Cx = \sqrt {2u + 2} \cr
& y = ux \to u = \frac{y}{x} \cr
& Cx = \sqrt {\frac{{2y}}{x} + 2} \cr
& {C^2}{x^2} = \frac{{2y}}{x} + 2 \cr
& {C^2}{x^3} = 2y + 2x \cr
& y = \frac{{{C_1}{x^3} - 2x}}{2} \cr} $$
