Here $p=\frac{dy}{dx}$. $$3p^2y^2-2xyp+4y^2-x^2=0 \\ p=\frac{2xy\pm\sqrt{4x^2y^2-4(3y^2)(4y^2-x^2)}}{6y^2} \\ p=\frac{2xy\pm2\sqrt{x^2y^2-(12y^4-3x^2y^2)}}{6y^2} \\ p=\frac{x\pm2\sqrt{x^2-3y^2}}{3y}$$
I am not able to proceed from here. What I have is the following. $$\frac{3ydy-xdx}{\sqrt{x^2-3y^2}}=\pm2ydx$$
For reference answer given in the book is
$$2x\pm(x^2-3y^2)^{1/2}=c$$
Sorry I made a mistake. I got it now. $$3y\frac{dy}{dx}-x=\pm\sqrt{x^2-3y^2} \\ \frac{3ydy-xdx}{\sqrt{x^2-3y^2}}=\pm dx \\ -\frac{1}{2}d(\sqrt{x^2-3y^2})=\pm dx \\ 2x\pm\sqrt{x^2-3y^2}=c$$