Answer
$y=ce^{-\frac{3}{2}\frac{x}{y}}$
Work Step by Step
Given:
$$(3x-2y)\frac{dy}{dx}=3y$$
Let $y=vx \rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Subtituting:
$$(3-2v)(v+x\frac{dv}{dx})=3v$$
$$x\frac{dv}{dx}=\frac{3v}{3-2v}-v$$
$$x\frac{dv}{dx}=\frac{3v-3v+2v^2}{3-2v}$$
$$x\frac{dv}{dx}=\frac{2v^2}{3-2v}$$
$$\frac{3-2v}{2v^2}dv=\frac{dx}{x}$$
Integrating left side:
$$\int \frac{3-2v}{2v^2}=\int \frac{dx}{x}+C$$
$$-\frac{3}{2}v^{-1}-\ln v=\ln|X| + C$$
where $c$ is a constant of integration.
$$\ln x + \ln v=-c-\frac{3}{2}v$$
$$\ln (xv)=-c-\frac{3}{2}v$$
Subtitute when $y=vx \rightarrow \ln y=-c-\frac{3}{2}\frac{x}{y}$
Solve for $y$
$$y=ce^{-\frac{3}{2}\frac{x}{y}}$$
