Stability of equilibrium in diff EQ symbiotic growth model 
Consider the following system, which is designed to model a symbiotic
  relationship between two species: $$\dot{x}(t) = x(\epsilon_1
 -\alpha_1 x + \beta_1 y)\\  \dot{y}(t) = y (\epsilon_2 + \beta_2x - \alpha_2y)$$
Here $\alpha_i,\, \beta_i, \, \epsilon_i\,$ are positive constants.
  
  
*
  
*Under what circumstances is there an equilibrium point with both    species present?
  
*What can you say about the stability of the equilibrium?
  

For an equilibrium point with both species present, we need $x(\epsilon_1 -\alpha_1 x + \beta_1 y)=y (\epsilon_2 + \beta_2x - \alpha_2y)=0$ and $x,y>0$. This gives us a linear system
$$
\left[  
\begin{matrix}
\alpha_1 & -\beta_1\\
-\beta_2 & \alpha_2
\end{matrix}
\right]
\left[
\begin{matrix}
x\\y
\end{matrix}
\right]
=
\left[
\begin{matrix}
\epsilon_1\\\epsilon_2
\end{matrix}
\right] 
$$
I claim that if $\alpha_1\alpha_2 - \beta_1\beta_2 = 0$, there are no solutions with $x,y>0$.  If $\alpha_1\alpha_2 - \beta_1\beta_2 \neq 0$, this has a unique solution
$$
x_{\text{eq}} = \frac{1}{\alpha_1\alpha_2 - \beta_1\beta_2}\left( \alpha_2 \epsilon_1 + \beta_1 \epsilon_2 \right)\\
y_{\text{eq}}= \frac{1}{\alpha_1\alpha_2 - \beta_1\beta_2}\left( \beta_2\epsilon_1 + \alpha_1 \epsilon_2 \right),
$$
and if we want $x,y>0$, we will need $\alpha_1\alpha_2 - \beta_1\beta_2>0$. So an equilibrium point with both species present exists iff $\alpha_1\alpha_2 - \beta_1\beta_2>0$.

Now, the stability of this point is the part I have a question about. I conjecture that any ball around the equilibrium point which stays in the first quadrant will be positive invariant, but I don't really know. I had no luck looking for a Lyapunov function. Any ideas?
 A: Since you mentioned looking for online reference, I suggest Scholarpedia article. Scholarpedia covers few topics compared to Wikipedia, but it seems to treat ODE well. 
In your case, let $(x^*,y^*)$ be the coexistence equilibrium. The Jacobian matrix at the equilibrium is easy to compute: using product rule, we see that the vanishing term needs to be differentiated. Result: 
$$\begin{pmatrix} -\alpha_1 x^* & \beta_1 x^* \\  \beta_2 y^* & -\alpha_2 y^* \end{pmatrix}$$
You don't actually need to find the eigenvalues of this matrix. In the 2-dimensional case, the trace and determinant tell  the story, as on the diagram below (credit: Douglas Hundley).
 
The diagram is for linear systems. For nonlinear systems, such as yours, the upper part of the vertical axis ("center") is the difficult case. In this case the nonlinear term can swing the behavior toward asymptotic stability or instability. Or keep it neutrally stable, as in the classical Lotka-Volterra example. 
A: It is a theorem that if the Jacobian of the function describing $\left[ \begin{matrix} x\\y \end{matrix}\right] ' $ is negative definite at the equilibrium point, then the equilibrium point is asymptotically stable. I claim the Jacobian of the function at $(x_{\text{eq}}, y_{\text{eq}})$ is 
$$
J(x_{\text{eq}}, y_{\text{eq}})=
$$
$$ 
\frac{1}{(\alpha_1\alpha_2-\beta_1\beta_2)^2}
\left[
\begin{matrix}
\epsilon_1(-\alpha_1\alpha_2) + \epsilon_2(-\alpha_1\beta_2) 
& \epsilon_1(\alpha_2\beta_1) + \epsilon_2({\beta_1}^2)
\\ \epsilon_1({\beta_2}^2) + \epsilon_2(\alpha_1\beta_2)
& \epsilon_1(-\alpha_1\beta_2) + \epsilon_2(-\alpha_1\alpha_2) 
\end{matrix}
\right].
$$
The trace of this matrix is clearly negative, so to show that it is negative definite, it suffices to show that the determinant is positive. I claim that the determinant is 
$$
{\epsilon_1}^2(\alpha_2\beta_2)(\alpha_1\alpha_2 - \beta_1\beta_2) + {\epsilon_2}^2(\alpha_1\beta_1)(\alpha_1\alpha_2 - \beta_1\beta_2) + \epsilon_1\epsilon_2(\alpha_1\alpha_2 + \beta_1\beta_2)(\alpha_1\alpha_2 - \beta_1\beta_2),
$$
and each term here is positive.
Edit: this answer is technically correct, but wastes a lot of effort. It's easier to simplify the Jacobian immediately, as user147263 did.
