Solving a first order ordinary differential equation with no initial conditions 
Solve the following differential equation. $$ \dfrac{dy}{dx} = \dfrac{3y - 3x}{2x+3y} $$


Answer:
Let $y = xv$. Then we have:
\begin{align*} 
\dfrac{dy}{dx} &= x \dfrac{dv}{dx} + v \\
v &= \dfrac{ y}{x} \\
\dfrac{dy}{dx} &=
 \dfrac{ 3\left( \dfrac{y}{x} \right)  - 3} {2+3\left( \dfrac{y}{x}\right) } \\ 
x \dfrac{dv}{dx} + v &= \dfrac{3v-3}{2+3v} \\
x \dfrac{dv}{dx}  &= \dfrac{3v-3 - v(2+3v)}{2+3v} \\
x \dfrac{dv}{dx}  &= \dfrac{3v-3 -2v - 3v^2}{2+3v} \\
x \dfrac{dv}{dx}  &= \dfrac{-3v^2+v-3}{3v+2} \\
\dfrac{dx}{x} &= \dfrac{ 3v + 2}{-3v^2 + v -3 } \,\,\, dv
\end{align*}
Using  an  online integral calculator, I find:
$$ \int \dfrac{ 3x+2 }{ -3x^2 + x - 3} \,\, dv =
 -\dfrac{ \ln( 3x^2 - x + 3 ) }{2} +
 \dfrac{4 \arctan\left( \dfrac{ 6x - 1 }{ \sqrt{35} } \right)  } { \sqrt{35} }.
  + C$$
Now we have:
\begin{align*} 
\ln{x} &=
 -\dfrac{ \ln( 3v^2 - v + 3 ) }{2} +
 \dfrac{4 \arctan\left( \dfrac{ 6v - 1 }{ \sqrt{35} } \right)  } { \sqrt{35} }.
 + C
\end{align*}
According to ChatGPT, the correct solution is:
$$ y = (2x+3y)(x+C) $$
Where did I go wrong?

Notice that
$$y=(2x+3y)(x+c)\iff y=\frac{2x(x+c)}{-3(x+c)+1},$$when $(x+c)\not=\frac{1}{3}$. Setting $c=0$ and find $dy/dx$ we have $$\frac{dy}{dx}=\frac{2(2-3x)x}{(3x-1)^2}\not=\frac{3(\frac{2x^2}{1-3x})-3x}{2x+3(\frac{2x^2}{1-3x})}.$$
So your function $y$ is not solution of the ode.
Instead, your first way is correct. Setting the substitution $y/x:=v$ we need to solve the ode
$$x\frac{dv}{dx}+v=\frac{3(v-1)}{3v+2}$$
Thus, we get by variable separation
$$-\sqrt{\frac{5}{7}}\arctan(\frac{6v-1}{\sqrt{35}})-\frac{1}{2}\ln|3v^2-v+3|=\ln|x|+c,$$with $c$ be an arbitrary constant, then substitution back give the implicit solution.
