Yes. "Equilibrium" solutions are "constant" solutions- solutions such that dx/dt= 0 and dy/dt= 0. Since dy/dt= -2x= 0, x= 0 which makes the second equation $\displaystyle dx/dt= -x+ y- y^2= y- y^2= y(1- y)= 0$ so that y= 0 or y= 1. There are a number of different ways of determining what "type" of equilibrium solution each of those is. I don't know which you are expected to use.