Non-linear system equilibrium points but critical point I have a question about critical points of nonlinear systems. So I am given the following prey-predator model
$$f(x,y) = x' = 10(1-\frac{x}{k})x-\frac{10xy}{1+x}$$
$$g(x,y) = y' = -y+\frac{10}{4}\frac{xy}{1+x}$$
It follows that x is the prey while y represents the predator population. So solving for this I found that the critical points are given by $(0,0),(\frac{2}{3},\frac{15k-10}{9k}),(k,0)$. I am asked to check for the stability of the equilibrium points for different values of $k$ where: 
$0<k<\frac{2}{3}$
$k = \frac{2}{3}$
$\frac{2}{3}<k<\frac{7}{3}$
$k^{*}\approx1.2667$
$k^{*} < k < \frac{7}{3}$
Most of the cases are find except for the point $k=\frac{2}{3}$ where I seem to be getting a jacobian that gives a $0$ eigenvalue. I find such results very odd because I thought that the local behavior at some points cannot results in a $det(A) = 0$. Moreover, I am told that I should find transitions in local behaviour of at least one equilibria at the following values: $K = 2/3$, $K = K^{*} ≃ 1.266$, $K = 7/3$, of which I seemed to get saddle points at critical points $(0,0)$, and $(K,0)$, while the last critical point seems to result in stable nodes except for $k=\frac{2}{3}$. 
So I am wondering if someone could clarify whether such results are possible for $k=\frac{2}{3}$. Additionally, I would also like to ask whether saddles are the only transitions and whether others (ie: repeated eigenvalue) also count. 
Edit: At $k=\frac{2}{3}$, it follows that the point critical points are (0,0),$(\frac{2}{3},y(\frac{2}{3}) = (\frac{2}{3},0)\text{ since } (\frac{15(\frac{2}{3})-10)}{9\frac{2}{3}} = 0)$ and like wise for the other critical point given by $(k,0) = (\frac{2}{3},0)$. So two of the critical points are the same. Evaluating the jacobian (sorry I didnt realize that I did not write the jacobian down). 
$$J(x,y)=\begin{pmatrix} f_x = 10(1-\frac{2x}{k}-\frac{y}{(1+x)^2}) & f_y = \frac{-10x}{1+x} \\ g_x=\frac{10y}{4}(\frac{1}{(1+x)^2} & g_y = -1 + \frac{10}{4}(\frac{x}{1+x}) \end{pmatrix}$$
Hence, $$J(\frac{2}{3},0)=\begin{pmatrix} -10 & -4 \\ 0 & 0 \end{pmatrix}$$
As a result, this gives a case where the Trace < $0$ but the Det = $0$ with $\lambda = 0$ and $\lambda = -10$
Thank you.
 A: The critical points are:
$$(x, y) = \left(\frac{2}{3},\frac{50 (3 k-2)}{9 k}\right),(k, 0),(0, 0)$$
At $k = \dfrac{2}{3}$, these reduce to:
$$(x, y) = \left(\frac{2}{3},0\right),(0, 0)$$
The Jacobian is given by:
$$J(x, y) = \begin{bmatrix} \dfrac{\partial f}{\partial x} & \dfrac{\partial f}{\partial y} \\ \dfrac{\partial g}{\partial x} & \dfrac{\partial g}{\partial y} \end{bmatrix} = \begin{bmatrix} -\dfrac{20 x}{k}-\dfrac{y}{(x+1)^2}+10 &  \dfrac{1}{x+1}-1\\ \dfrac{5 y}{2 (x+1)^2} &  \dfrac{5 x}{2 (x+1)}-1 \end{bmatrix} $$
At $k = \dfrac{2}{3}$, the Jacobian is:
$$J(x, y) = \begin{bmatrix} -\dfrac{y}{(x+1)^2}-30 x+10 &  \dfrac{1}{x+1}-1\\ \dfrac{5 y}{2 (x+1)^2} &  \dfrac{5 x}{2 (x+1)}-1 \end{bmatrix} $$
At the critical point $(0, 0)$, when  $k = \dfrac{2}{3}$, we have the Jacobian:
$$\begin{bmatrix} 10 &  0\\ 0 &  -1 \end{bmatrix} $$
At the critical point $\left(\dfrac{2}{3}, 0\right)$, when  $k = \dfrac{2}{3}$, we have the Jacobian:
$$\begin{bmatrix} -10 &  -\dfrac{2}{5}\\ 0 & 0 \end{bmatrix} $$
A phase portrait at $k = \dfrac{2}{3}$ shows:

It appeared that this and the Jacobian at that point was what was giving trouble, so I stopped there regarding everything else as you claim you did not have issue with those.
