Solving for \(y\) gives\begin{align} y & =\ln \left ( \cos p\right ) +p\tan p\tag {1}\\ y & =xf+g\nonumber \\ & =g\tag {1A}\end{align}
Where \(f=0\) and \(g\left ( p\right ) =\ln \left ( \cos p\right ) +p\tan p\). Important note: This ode has \(f=0\) which is strictly speaking is not of the form \(y=xf\left ( p\right ) +g\left ( p\right ) \). But Maple says this is d’Alembert. This is why it is included. I should make special case d’Almbert classification to handle this special case.
Taking derivative of (1A) w.r.t. \(x\) gives\begin{align} p & =\frac {dg}{dp}\frac {dp}{dx}\nonumber \\ p & =\left ( -\frac {\sin p}{\cos p}+\tan p+p\left ( 1+\tan ^{2}p\right ) \right ) \frac {dp}{dx}\nonumber \\ p & =\left ( -\tan p+\tan p+p\left ( 1+\tan ^{2}p\right ) \right ) \frac {dp}{dx}\nonumber \\ p & =p\left ( 1+\tan ^{2}p\right ) \frac {dp}{dx}\nonumber \\ 1 & =\left ( 1+\tan ^{2}p\right ) \frac {dp}{dx}\end{align}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which does not result in solution.
The general solution is found by finding \(p\) from (2). Since (2) is not linear in \(p\), then inversion is needed. Writing (1) as\begin{align*} \frac {dx}{dp} & =1+\tan ^{2}p\\ dx & =\left ( 1+\tan ^{2}p\right ) dp \end{align*}
Integrating gives\begin{align*} x & =\tan p+c\\ p & =\arctan \left ( x-c\right ) \end{align*}
Substituting the above in (1) gives the solution\begin{align*} y & =\ln \left ( \cos p\right ) +p\tan p\\ & =\ln \left ( \cos \left ( \arctan \left ( x-c\right ) \right ) \right ) +\left ( \arctan \left ( x-c\right ) \right ) \tan \left ( \arctan \left ( x-c\right ) \right ) \\ & =\ln \left ( \cos \left ( \arctan \left ( x-c\right ) \right ) \right ) +\left ( x-c\right ) \arctan \left ( x-c\right ) \end{align*}
This ode also have solution \(y=0\).