Stability analysis using the Jacobian I must find the steady states, the Jacobian, and the stability of each point. 
$x' =x^2 - y^2$
and $y' = x(1-y)$
Simply solving for when these equations equal zero, I found that $y=0,1$, after getting $y-yy=0$, as must x, as $x^2=y^2$. 
Thus I believe the steady states to be $(0,0)$ and $(1,1)$ but I don't understand the Jacobian matrix. I think it is:
$
  \left[ {\begin{array}{cc}
   2x-y^2 & x^2-2y \\
   1-y & x-1 \\
  \end{array} } \right]
$
So for the state (0,0):
$
  \left[ {\begin{array}{cc}
   0 & 0 \\
   1 & 1 \\
  \end{array} } \right]
$
So, the trace of the matrix is $1$ and the determinant is $0$. Is this neutral center?
For (1,1):
$
  \left[ {\begin{array}{cc}
   0 & 0 \\
   0 & 0\\
  \end{array} } \right]
$
So the trace and the determinant are both $0$ (how do we characterize this)
Any help would be appreciated, as would work checking, am new to differential equations. 
 A: Hint: You are missing an equilibrium point.
$$\dot{x}=0=x^2-y^2 \implies x=\pm y$$
$$\dot{y}=0=x(1-y) \implies x=0 \text{ or } y=1$$
Using $x=0$ for the first equation we see that $y=0$, which implies the first equilibrium at $(0,0)$.
Using $y=1$ for the first equation we see that $0=x^2-1\implies x=\pm 1$, which implies the second and third equilibrium points $(1,1)$ and $(-1,1)$.
Using $x=\pm y$ for the second equation we obtain:
$$0=\pm y (1-y) \implies y = 0 (x=\pm 0) \text { and } y = 1 (x=\pm1).$$
Hence, in total we only have three equilibria: $(0,0)$, $(1,1)$ and $(-1,1)$.
The Jacobian of your system is given by:
$$J =\begin{bmatrix}2x & -2y \\ 1-y & -x\end{bmatrix}$$
Now, your task is to evaluate the Jacobian at the equilibrium points.
You can then determine the eigenvalues of the system and discriminate three cases:


*

*All eigenvalues $\lambda_i$ have real part strictly smaller than zero ($\operatorname{Re}(\lambda_i)<0$ for all $i$) -> You can conclude asymptotic stability for the nonlinear equilibrium point.

*You have at least one eigenvalue with real part strictly greater than zero ($\exists \lambda_i \text{ such that } \operatorname{Re}(\lambda_i)>0$)-> You can conclude instability for the nonlinear equilibrium point. The equilibrium points $(1,1)$ and $(-1,1)$ do belong in this category.

*All eigenvalues have a real part which is smaller or equal to zero and you have at least one eigenvalue with a real part of zero -> Linearization does not yield any stability information (this case is sometimes referred as the critical case). The equilibrium point $(0,0)$ belongs in this category. A stability analysis can be carried out if one is able to find a proper Lyapunov function.

