How to proceed to find the general and singular solution of the equation
3xy=2px^{2}-2p^{2}, p=dy/dx
When you write $\displaystyle \displaystyle \begin{align*} p^2 \end{align*}$, are you meaning $\displaystyle \displaystyle \begin{align*} \left( \frac{dy}{dx} \right) ^2 \end{align*}$ or $\displaystyle \displaystyle \begin{align*} \frac{d^2y}{dx^2} \end{align*}$?
I would probably try something like this:
$\displaystyle \displaystyle \begin{align*} 3x\,y &= 2x^2\,\frac{dy}{dx} - 2\left( \frac{dy}{dx} \right) ^2 \\ 2 \left( \frac{dy}{dx} \right) ^2 - 2x^2\,\frac{dy}{dx} + 3x\,y &= 0 \\ \left( \frac{dy}{dx} \right) ^2 - x^2\,\frac{dy}{dx} + \frac{3}{2}x\,y &= 0 \\ \left( \frac{dy}{dx} \right) ^2 - x^2 \, \frac{dy}{dx} + \left( -\frac{1}{2}x^2 \right) ^2 - \left( -\frac{1}{2}x^2 \right) ^2 + \frac{3}{2}x\,y &= 0 \\ \left( \frac{dy}{dx} - \frac{1}{2}x^2 \right) ^2 - \frac{1}{4}x^2 + \frac{3}{2}x\,y &= 0 \\ \left( \frac{dy}{dx} - \frac{1}{2}x^2 \right) ^2 &= \frac{x^2 - 6x\,y}{4} \\ \frac{dy}{dx} - \frac{1}{2}x^2 &= \frac{\pm \sqrt{x^2 - 6x\,y}}{2} \\ \frac{dy}{dx} &= \frac{x^2 \pm \sqrt{x^2 - 6x\,y}}{2} \end{align*}$
I don't know if this looks any easier to solve...