Stability of the given nonlinear system 
Consider the system 
  $$x'=-x+y^2,~~~~y'=x-2y+y^2.$$

(i). Show that $x=y$ is an orbit and is (for $x>0$) the unstable manifold through one of the critical points.
(ii). Show that $|x-y| \rightarrow 0$ as $t \rightarrow \infty$ for all the other orbits.
So by setting $x'=0$ and $y'=0,$ two critical points are found: $(0,0)$ and (1,1). I don't know how to go by and conclude what's being asked. Any help is much appreciated.
 A: For $x=y$ you get $$x'=y'=-x+x^2.$$ This readily implies that the line $x=y$ is a union of orbits. Why? (But it is not an orbit.)
The remainder of the problem is just a matter of checking what are the directions on the line $x=y$.
A: Another (partly) approach is doing what you have begun and looking for the equilibria
$$x=y^2\Rightarrow 2y^2-2y=2y(y-1)=0 \Rightarrow y_1=1,y_2=0 \Rightarrow (1,1),(0,0)$$
 and then looking at their stability and determining their stabl/unstable/center eigenspaces. 
$$Df(x,y)=\begin{pmatrix} -1 & 2y \\ 1 & -2+2y \end{pmatrix}$$
1) Let's look at $(0,0)$:
$$Df(0,0)=\begin{pmatrix} -1 & 0 \\ 1 & -2 \end{pmatrix}$$
Eigenvalues $-1,-2$ with the stable eigenspace $E^s=\langle\binom{1}{1},\binom{0}{1}\rangle$ (these are the eigenvectors to the eigenvalues with negative real part).
2) Let's look at $(1,1)$: $$Df(1,1)=\begin{pmatrix} -1 & 2 \\ 1 & 0 \end{pmatrix}$$
Eigenvalues $-2,1$ with the stable eigenspace $E^s=\langle\binom{-2}{1} \rangle$ and unstable eigenspace $E^u=\langle\binom{1}{1} \rangle$. That means $x=y$ indeed describes the unstable eigenspace of the equilibrium $(1,1)$.
