I have the following system:
$$x'=x(1-x)-xy$$ $$y'=y(1-\frac{y}{x})$$
I am trying to linearize it, and from my understanding, if the equilibrium points are hyperbolic, that is, its eigenvalues all have a non-zero real part, I can drop the higher order terms. If I compute $DF_{X_0}$, I get:
$$\begin{pmatrix} 1-2x-y & -x \\ \frac{y^2}{x^2} & 1-\frac{2y}{x} \end{pmatrix}$$
When I drop the higher order terms, I am left with:
$$\begin{pmatrix} 1-2x-y & -x \\ 0 & 1 \end{pmatrix}$$
Am I linearizing this correctly?
No, you have misunderstood this, you should not drop higher order terms in the Jacobian like you are doing! What you should to is to substitute the equilibrium that you want to investigate into the Jacobian. That will give you a constant matrix containing the coefficients of the first-order terms in the Taylor expansion around that point, so the “dropping of terms” is automatically taken care of.