\(\left ( y^{\prime }\right ) ^{2}-1-x-y=0\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives\begin {align} y & =-x+\left ( p^{2}-1\right ) \tag {1}\\ & =xf+g\nonumber \end {align}
Hence \(f=-1,g\left ( p\right ) =\left ( p^{2}-1\right ) \). 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 ) \nonumber \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\nonumber \\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \tag {2} \end {align}
Using \(f=-1,g=\left ( p^{2}-1\right ) \) the above simplifies to\begin {equation} p+1=2p\frac {dp}{dx} \tag {2A} \end {equation} The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=-1\). Substituting this in (1) gives singular solution as\begin {equation} y\left ( x\right ) =-x \tag {3} \end {equation} The general solution is found by finding \(p\) from (2A). No need here to do the inversion as (2) is separable already. Solving (2) gives\begin {align*} p & =-\operatorname *{LambertW}\left ( -e^{-\frac {x}{2}-1+\frac {c_{2}}{2}}\right ) -1\\ & =-\operatorname *{LambertW}\left ( -c_{1}e^{-\frac {x}{2}-1}\right ) -1 \end {align*}
Substituting the above in (1) gives the general solution\begin {align} y\left ( x\right ) & =-x+\left ( p^{2}-1\right ) \nonumber \\ y\left ( x\right ) & =-x+\left ( -\operatorname *{LambertW}\left ( -c_{1}e^{-\frac {x}{2}-1}\right ) -1\right ) ^{2}-1 \tag {4} \end {align}
Note however that when \(c_{1}=0\) then the general solution becomes \(y\left ( x\right ) =-x\). Hence (3) is a particular solution and not a singular solution. Solution (4) is therefore the only solution.