If you are interested I have developed a symbolic package in Mathematica for obtaining the symmetries of a DE. You can learn more about it here. If you are interested I can send you an updated version of the package via email.
This specific ODE is a first order one so don't expect to find a finite Lie group without making an ansantz in the form of the symmetries. For example, let your symmetries be represented by the infinitesimal generator $\mathcal X=\xi(x,y)\partial_x+\eta(x,y)\partial_y$ as a Lie algebra and assume that the coefficients $\xi,\eta$ are polynomials of order up to three with respect to $x,y$ then we have the symmetries: \begin{array}{l} (-2-x+2 y)\partial _x \\ (2+x)\partial _x+y\partial _y \\ 2 x y\partial _y+x (6+3 x-2 y)\partial _x \\ 2 x y^2\partial _y+x (6+3 x-2 y) y\partial _x \\ 2 y^3\partial _y+(6+3 x-2 y) y^2\partial _x \\ 2 x y (-6+3 x+2 y)\partial _y+x \left(9 x^2-4 (-3+y)^2\right)\partial _x \\ 2 y^2 (3+y)\partial _y+\left(3 x \left(3+3 y+y^2\right)-2 \left(-9+y^3\right)\right)\partial _x \\ \end{array}
You can see that the symmetry found by Mariano Suárez-Alvarez is the second one from the above list.
From that point by exponentiation you can obtain the respective local Lie group trasformation. For example, for the second symmetry we have the ode's \begin{array}{l} \bar x^\prime(\epsilon) = \bar x(\epsilon)+2,\, \bar x(0)=x\\ \bar y^\prime(\epsilon) = \bar y,\, \bar y(0)=y \end{array} by solving this system of ODEs you get \begin{array}{l} \bar x(\epsilon) =e^\epsilon(x+2)-2=\lambda (x+2)-2\\ \bar y(\epsilon) =e^\epsilon y=\lambda y \end{array}
Now, to obtain a solution for the ODE you can you use one of the symmetries to get a canonical set of coordinates. For example, from the second one the change of coordinates $ X=\frac{y}{x+2},\, Y=\ln\left|y\right| $ will turn the symmetry to the canonical form $\partial_Y$. By applying this change of variables to the ODE you get the ODE: $ Y^\prime = \frac{1}{X(X-\frac{1}{2})}. $ After solving it and returning back to the original variables you have: $ y^3=c(x+2-2y)^2. $ Bottomline: symmetries encompass most of the empirical methods we know for solving ODEs and PDEs, but the advantage of Lie's method is that they can be obtained in a systematic and algorithmic way.