\begin {align*} y & =x+\left ( y^{\prime }\right ) ^{2}\left ( 1-\frac {2}{3}y^{\prime }\right ) \\ & =x+p^{2}\left ( 1-\frac {2}{3}p\right ) \end {align*}
Where \(f=1,g=p^{2}\left ( 1-\frac {2}{3}p\right ) \). Taking derivative w.r.t. \(x\) gives\begin {align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \end {align*}
Using values for \(f,g\) the above simplifies to\begin {equation} p-1=\left ( 2p-2p^{2}\right ) \frac {dp}{dx} \tag {2A} \end {equation} The singular solution is when \(\frac {dp}{dx}=0\) which results in \(p=1\). Substituting this in (1) gives\begin {align*} y & =x-\left ( 1-\frac {2}{3}\right ) \\ & =x+\frac {1}{3} \end {align*}
The general solution is when \(\frac {dp}{dx}\neq 0\). Then (2A) is now separable. Solving for \(p\) gives\begin {align*} p & =-\sqrt {c_{1}-x}\\ p & =\sqrt {c_{1}-x} \end {align*}
Substituting each one of the above solutions of \(p\) in (1) gives \begin {align*} y_{1} & =x+\left ( p^{2}-\frac {2}{3}p^{3}\right ) \\ & =x+\left ( \left ( -\sqrt {c_{1}-x}\right ) ^{2}-\frac {2}{3}\left ( -\sqrt {c_{1}-x}\right ) ^{3}\right ) \\ & =x+\left ( c_{1}-x+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\right ) \\ & =c_{1}+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}} \end {align*}
And\begin {align*} y_{2} & =x+\left ( p^{2}-\frac {2}{3}p^{3}\right ) \\ & =x+\left ( \left ( \sqrt {c_{1}-x}\right ) ^{2}-\frac {2}{3}\left ( \sqrt {c_{1}-x}\right ) ^{3}\right ) \\ & =x+\left ( c_{1}-x-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\right ) \\ & =c_{1}-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}} \end {align*}
Therefore the solutions are\begin {align*} y & =x+\frac {1}{3}\\ y & =c_{1}+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\\ y & =c_{1}-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}} \end {align*}