Only **linear** transformations have eigenvalues (which is why you learn about them in **Linear** Algebra). What this problem is asking you to do is to find eigenvalues for the *linearized* systems, linearized close to the steady state.

Yes, (x, y)= (0, 0) is a steady state and you "linearize" a system by arguing that higher powers of small numbers can be ignored. Around (0, 0) we can linearize by ignoring higher powers of x and y (which includes products of x and y). The original system,

\(\displaystyle \frac{dx}{dt}= x(1- 2y- x)= x- 2xy- x^2\)

\(\displaystyle \frac{dy}{dt}= y- x\)

linearized around (0, 0) is

\(\displaystyle \frac{dx}{dt}= x\)

\(\displaystyle \frac{dy}{dt}= y - x\).

The vector form of that could be written

\(\displaystyle \frac{d}{dt}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix} 1 & 0 \\ -1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}\).

The "eigenvalues" referred to are the eigenvalues of the matrix \(\displaystyle \begin{bmatrix} 1 & 0 \\ -1 & 1\end{bmatrix}\)

To linearize around the other steady state, (x, y)= (1/3, 1/3), change variables: let u= x- 1/3 and v= y- 1/3 so that (x, y)= (1/3, 1/3) corresponds to (u, v)= (0, 0). Then x= u+ 1/3 and y= v+ 1/3 so the equations become

\(\displaystyle \frac{dx}{dt}= \frac{du}{dt}= (u+ 1/3)(1- 2(v+1/3)- u- 1/3)= (u+1/3)(1- 2v- u)\)\(\displaystyle = u- 2uv- u^2+ 1/3- (2/3)v- (1/3)u= (2/3)u- (2/3)v- 2uv- u^2)\) and

\(\displaystyle \frac{dy}{dt}= \frac{dv}{dt}= v+ 1/3- u- 1/3= v- u\)

Now, (x, y) close to (1/3, 1/3) means that (u,v) is close to (0, 0) so we can "linearize" by ignoring higher powers of u and v:

\(\displaystyle \frac{du}{dt}= (2/3)u- (2/3)v\) and

\(\displaystyle \frac{dv}{dt}= v- u\)

That can be written as

\(\displaystyle \frac{d}{dt}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}\frac{2}{3} & -\frac{2}{3} \\ -1 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}\)

The "eigenvalues" are the eigenvalues of the matrix \(\displaystyle \begin{bmatrix}\frac{2}{3} & -\frac{2}{3} \\ -1 & 1 \end{bmatrix}\)