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 $(x_0, y_0) = (\frac{\gamma}{\gamma + \alpha}, \frac{\gamma}{\gamma + \alpha})$. I have tried to linearise the system by pluging $x = x_0 + \epsilon_x$ and $y = y_0 + \epsilon_y$ in the system and managed to linearise the ODE for $\dot \epsilon_x$, but for $\dot \epsilon_y$ I got $$\dot \epsilon_y = (a+\epsilon_y)(1- \frac{a+\epsilon_y}{a+\epsilon_x}),$$ where $a = x_0 = y_0$.
Question:
How can I linearise the equation for $\dot \epsilon_y$?
I have tried expanding it around $(0,0)$ as a Taylor series, but all the terms of order smaller than $2$ vanish, so I need to deal with terms of order $2$; but that does not help me solving the problem. After all, the reason why we do linearisation is that we can find the eigenvalues of the linearised system to determine the stability of the fixed points of the non-linearised system.
If you expand the second factor arithmetically, you get $$ \dot ϵ_y=(a+ϵ_y)\frac{ϵ_x-ϵ_y}{a+ϵ_x}. $$ As the difference $ϵ_x-ϵ_y$ is already first-order small, all the other factors need only be expanded to first order. Assuming $a\ne 0$, only the constant term $a$ remains and cancels from the expression. Thus the linearization is $$ \dot ϵ_y=ϵ_x-ϵ_y. $$