You could always solve the DE and consider the solution as t--> oo ......

Alternativley, if the long term behaviour is that y --> constant, then dy/dt --> 0 => y^2 - 4y + 2 --> 0 .....

Since y(0) = 0 is under y = 2 - sqrt{2}, it looks to me that y --> 2 - sqrt{2} for this initial condition.

Since y(0) = 1 is above y = 2 - sqrt{2} and below y = 2 + sqrt{2}, it looks to me like y --> 2 + sqrt{2} for this initial condition.