<s>[INST] Proving a solution of a Bernoulli type equation Prove that
\begin{equation}
y(x) = \sqrt{\dfrac{3x}{2x + 3c}}
\end{equation}
is a solution of
\begin{equation}
\dfrac{dy}{dx} + \dfrac{y}{2x} = -\frac{y^3}{3x}
\end{equation}
All the math to resolve this differential equation is already done. The exercise simply asks to prove the solution.
I start by pointing out that it has the form
\begin{equation}
\dfrac{dy}{dx} + P(x)y = Q(x)y^3
\end{equation}
where
\begin{equation}
P(x) = \dfrac{1}{2x}, \qquad Q(x) = -\frac{1}{3x}
\end{equation}
Rewriting y(x) as
\begin{equation}
y(x) = (3x)^{\frac{1}{2}} (2x + 3c)^{-\frac{1}{2}}
\end{equation}
Getting rid of that square root, I'll need it later on to simplify things
\begin{equation}
[y(x)]^2 = 3x(2x + 3c)^{-1}
\end{equation}
Calculating dy/dx
\begin{align}
\dfrac{dy}{dx} &= \frac{1}{2}(3x)^{-\frac{1}{2}}(3)(2x + 3c)^{-\frac{1}{2}} + \left(-\dfrac{1}{2}\right)(2x + 3c)^{-\frac{3}{2}}(2)(3x)^{\frac{1}{2}} \\
&= \frac{3}{2}(3x)^{-\frac{1}{2}}(2x + 3c)^{-\frac{1}{2}} - (3x)^{\frac{1}{2}}(2x + 3c)^{-\frac{3}{2}} \\
&= (3x)^{\frac{1}{2}}(2x + 3c)^{-\frac{1}{2}} \left[\dfrac{3}{2}(3x)^{-1} - (2x + 3c)^{-1}\right] \\
&= y \left[\dfrac{3}{2}(3x)^{-1} - (2x + 3c)^{-1}\right] \\
&= \dfrac{y}{2x} - y(2x + 3c)^{-1} \\
&= \dfrac{y}{2x} - y\left(\dfrac{y^2}{3x}\right) \\
&= \dfrac{y}{2x} - \dfrac{y^3}{3x}
\end{align}
Finally
\begin{align}
\dfrac{dy}{dx} + P(x)y &= \dfrac{y}{2x} - \dfrac{y^3}{3x} + \dfrac{y}{2x} \\
&= \dfrac{y}{x} - \dfrac{y^3}{3x}
\end{align}
which obviously isn't the same as equation 2. I don't know where I screwed up.
 [/INST] You could probably get an easier calculation by going over the logarithmic derivative,
$$
\log(y(x))=\frac12(\log(3x)-\log(2x+3c))
\\~\\
\implies
\frac{y'(x)}{y(x)}=\frac12\left(\frac1x-\frac2{2x+3c}\right)
=\frac1{2x}-\frac1{3x}y(x)^2
$$
The last step is obtained by doing the minimum to eliminate the constant $c$.
As you can see, this again confirms your result.

As to the original equation itself, it solves as
$$
(y^{-2})'=-2y^{-3}y'=\frac{y^{-2}}x+\frac2{3x}\\
\left(\frac{y^{-2}}x\right)'=\frac2{3x^2}\implies \frac{y^{-2}}x=-\frac2{3x}+c\\
y(x)=\pm\sqrt{\frac3{3cx-2}}
$$
which is quite different from your solution formula.
</s>