This can be written: (1-x)dy=y^2dx; separating the variables: y^-2dy=dx/(1-x).
Integrate both sides: -y^-1=-ln|1-x|-a where a is a constant; y^-1=1/y=ln|1-x|+a=ln|1-x|+ln(A) where A=e^a, so 1/y = ln|A(1-x)| or y=1/ln|A(1-x)| or yln|A(1-x)|=1.
This can be written e^(1/y)=A(1-x).