Applying the 'standard procedure' the substitution $\displaystyle w=\frac{1}{y}$ leads to the DE...
$\displaystyle w^{'}= - 1+2 w x -2 w^{2}$ (1)
... which is known as 'Riccati Equation'. The solution of this type of equation is usually an hard task but, may be, not impossible in this case. The question requires further investigation...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
The original DE...
$\displaystyle y^{'}=2-2 x y +y^{2}$ (1)
... is itself a 'Riccati Equation' so that a substitution of the type $\displaystyle w=\frac{1}{y}$ is useless. More promising is the substiution $\displaystyle w=y-x$ after that the (1) becomes...
$\displaystyle w^{'}=1-x^{2} + w^{2}$ (2)
The (2) is also a 'Riccati Equation' but, very important detail, we can verify that $\displaystyle w_{1}=x$ is solution of (2) and that allows You to find the general solution as follows. Setting $\displaystyle w=w_{1}+u=x+u$ the (2) becomes...
$\displaystyle u^{'}=2 u x + u^{2}$ (3)
... which is a 'Bernoulli Equation', the solution of which is left to You...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$