I have the following dynamical system
$$\frac{d x}{d \tau}=\gamma x(1-x)-\alpha x y$$ $$\frac{d y}{d \tau}=y\left(1-\frac{y}{x}\right),$$ where $\gamma$ and $\alpha$ are constant parameters. I am trying to figure out the stability of the fixed point $(x_0, y_0) = (\frac{\gamma}{\gamma + \alpha}, \frac{\gamma}{\gamma + \alpha})$
When I perturb the system around the fixed point such that $x = x_0+ \epsilon_x$ and $y = y_0 + \epsilon_y$, and linearise the system, I get the following jacobian matrix $$J = \begin{pmatrix} \frac{\alpha \gamma}{\gamma + \alpha} & -\frac{\alpha \gamma}{\gamma + \alpha}\\ 1 & -1 \end{pmatrix}.$$
Since one of the eigenvalues of this is zero, the linearisation fail, i.e we cannot determine the stability of the fixed point of the non-linear system from the stability of the fixed point of the linear system.
This means that I have to solve the nonlinear system of equation for the perturbations in order to determine the stability of the fixed point,
$$\dot \epsilon_x = \left(\frac{\alpha \gamma}{\gamma + \alpha}\right)\epsilon_x - \left(\frac{\alpha \gamma}{\gamma + \alpha}\right)\epsilon_y - \gamma \epsilon_x^2 - \alpha \epsilon_x \epsilon_y$$ $$\dot \epsilon_y = [ \left(\frac{\alpha \gamma}{\gamma + \alpha}\right) + \epsilon_y][1-\frac{ \left(\frac{\alpha \gamma}{\gamma + \alpha}\right) + \epsilon_y}{ \left(\frac{\alpha \gamma}{\gamma + \alpha}\right) + \epsilon_x}].$$
Question:
How to find the stability of this fixed point, given that the one of the eigenvalues of the linearised system is zero and the other it negative?
Some comments:
The equilibrium points obtained as solutions for
$$ \cases{ \gamma x(1-x)-\alpha x y = 0\\ y(1-\frac yx)=0} $$
are
$$ \left(1,0\right),\ \ \ \left(\frac{\gamma}{\alpha+\gamma},\frac{\gamma}{\alpha+\gamma}\right) $$
The jacobian gives
$$ J = \left( \begin{array}{cc} \gamma (1-x)-\gamma x-\alpha y & -\alpha x \\ \frac{y^2}{x^2} & 1-\frac{2 y}{x} \\ \end{array} \right) $$
and evaluated at the equilibrium points give
$$ J_1 = \left( \begin{array}{cc} -\gamma & -\alpha \\ 0 & 1 \\ \end{array} \right),\ \ \ J_2 = \left( \begin{array}{cc} -\frac{\gamma ^2}{\alpha +\gamma } & -\frac{\alpha \gamma }{\alpha +\gamma } \\ 1 & -1 \\ \end{array} \right) $$
with the correspoding eigenvalues
$$ (1,\ -\gamma),\ \ \ \ \left(-\frac{\sqrt{\left(\alpha +\gamma ^2+\gamma \right)^2-4 \gamma (\alpha +\gamma )^2}+\alpha +\gamma ^2+\gamma }{2 (\alpha +\gamma )},-\frac{-\sqrt{\left(\alpha +\gamma ^2+\gamma \right)^2-4 \gamma (\alpha +\gamma )^2}+\alpha +\gamma ^2+\gamma }{2 (\alpha +\gamma )}\right) $$
so normally, no zero eigenvalues.