\[ \left ( y^{\prime }\right ) ^{2}x+y^{\prime }y\ln y-y^{2}\left ( \ln y\right ) ^{4}=0 \] Applying p-discriminant method gives\begin {align*} F & =\left ( y^{\prime }\right ) ^{2}x+y^{\prime }y\ln y-y^{2}\left ( \ln y\right ) ^{4}=0\\ \frac {\partial F}{\partial y^{\prime }} & =2y^{\prime }x+y\ln y=0 \end {align*}
We first check that \(\frac {\partial F}{\partial y}\neq 0\). Now we apply p-discriminant. Eliminating \(y^{\prime }\). Second equation gives \(y^{\prime }=-\frac {y}{2x}\ln y\). Substituting into the first equation gives\begin {align*} \left ( -\frac {y}{2x}\ln y\right ) ^{2}x+\left ( -\frac {y}{2x}\ln y\right ) y\ln y-y^{2}\left ( \ln y\right ) ^{4} & =0\\ \frac {y^{2}}{4x}\left ( \ln y\right ) ^{2}-\frac {y^{2}}{2x}\left ( \ln y\right ) ^{2}-y^{2}\left ( \ln y\right ) ^{4} & =0\\ y^{2}\ln \left ( y\right ) ^{2}\left ( \frac {1}{4x}-\frac {1}{2x}-\left ( \ln y\right ) ^{2}\right ) & =0 \end {align*}
Hence we obtain the solutions\begin {align*} y & =0\\ y & =\infty \\ \frac {1}{4x}-\frac {1}{2x}-\left ( \ln y\right ) ^{2} & =0 \end {align*}
Or\begin {align} y & =0\tag {1}\\ y & =\infty \tag {2}\\ 1+4x\left ( \ln y\right ) ^{2} & =0 \tag {3} \end {align}
The solution \(y=0\) does not satisfy the ode. Same for \(y=\infty \). The solution \(1+4x\left ( \ln y\right ) ^{2}=0\) gives \(y=\left \{ \begin {array} [c]{c}e^{\frac {-i}{2\sqrt {x}}}\\ e^{\frac {i}{2\sqrt {x}}}\end {array} \right . \) and only the first one satisfies the ode and for negative \(x\) only. The primitive can be found to be \[ \Psi \left ( x,y,c\right ) =y-e^{\frac {c}{c^{2}-x}}=0 \] Now we have to eliminate \(c\) using the c-discriminant method\begin {align*} \Psi \left ( x,y,c\right ) & =y-e^{\frac {c}{c^{2}-x}}=0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =\left ( \frac {1}{c^{2}-x}-\frac {2c^{2}}{\left ( c^{2}-x\right ) ^{2}}\right ) e^{\frac {c}{c^{2}-x}}=0 \end {align*}
Second equation gives \(\frac {1}{c^{2}-x}-\frac {2c^{2}}{\left ( c^{2}-x\right ) ^{2}}=0\). Hence \(c=\pm \sqrt {-x}\). Substituting \(\sqrt {-x}\) in first equation gives\begin {align} y-e^{\frac {\sqrt {-x}}{-x-x}} & =0\nonumber \\ y & =e^{\frac {\sqrt {-x}}{-2x}}\nonumber \\ \ln y & =\frac {\sqrt {-x}}{-2x}\nonumber \\ \left ( \ln y\right ) ^{2} & =\frac {-x}{4x^{2}}\nonumber \\ 4x\left ( \ln y\right ) ^{2}+1 & =0\nonumber \\ y_{s} & =\left \{ \begin {array} [c]{c}e^{\frac {-i}{2\sqrt {x}}}\\ e^{\frac {i}{2\sqrt {x}}}\end {array} \right . \tag {4} \end {align}
Substituting \(-\sqrt {-x}\) in first equation gives same solution. But only the first one satisfies the ode for \(x<0\). This is the same as obtained using p-discriminant, hence it is a singular solution.
There is another singular solution \(y_{s}=1\). But now I do not know how to find this. I need to look more into this. See paper by C.N. SRINIVASINGAR, example 4. The following plot shows the singular solution above as the envelope of the family of general solution plotted using different values of \(c\). Added also \(y_{s}=1\).