This is homogeneous ODE of Class A of form \(y^{\prime }=F\left ( \frac {y}{x}\right ) \), hence from the lookup table \begin {align*} \xi & =x\\ \eta & =y \end {align*}
Canonical coordinates \(\left ( R,S\right ) \) are found similar to the above which gives\begin {align*} R & =\frac {y}{x}\\ S & =\ln y \end {align*}
What is left is to find \(\frac {dS}{dR}\). This is given by\[ \frac {dS}{dR}=G\left ( R\right ) \] Which is the same as above\[ \frac {dS}{dR}=\frac {\frac {dy}{dx}}{-R^{2}+R\frac {dy}{dx}}\] But in this problem, the only difference is that \(\frac {dy}{dx}=\frac {-3+\frac {y}{x}}{-1-\frac {y}{x}}=\frac {-3+R}{-1-R}\), hence\begin {align*} \frac {dS}{dR} & =\frac {\frac {-3+R}{-1-R}}{-R^{2}+R\left ( \frac {-3+R}{-1-R}\right ) }\\ & =\frac {1}{R}\frac {R-3}{R^{2}+2R-3} \end {align*}
Which is a quadrature. In Lie method, for first order ode, we always obtain \(\frac {dS}{dR}=G\left ( R\right ) \). Integrating the above gives\begin {align*} \int dS & =\int \frac {1}{R}\left ( \frac {R-3}{R^{2}+2R-3}\right ) dR\\ S & =\ln \left ( R\right ) -\frac {1}{2}\ln \left ( R+3\right ) -\frac {1}{2}\ln \left ( R-1\right ) +c_{1} \end {align*}
Final step is to replace \(R,S\) back with \(x,y\) which gives\[ \ln y=\ln \left ( \frac {y}{x}\right ) -\frac {1}{2}\ln \left ( \frac {y}{x}+3\right ) -\frac {1}{2}\ln \left ( \frac {y}{x}-1\right ) +c_{1}\] This can be solved for \(y\) if an explicit solution is needed.