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Victor Ivrii:
If you discovered that at the critical point eigenvalues for the linearized system are purely imaginary, for non-linear system it may be either center, or stable focal point, or unstable focal point. However for this system we can exclude focal points. Why?

Victor Ivrii:
One needs to remember, that there are two major cases: General and Integrable. Below subscripts denote partial derivatives (as usual)

In the general case
$$\begin{pmatrix}x\\y\end{pmatrix}'=\begin{pmatrix}F(x,y)\\G(x,y)\end{pmatrix}\tag{1}$$
stationary points are those $(\bar{x},\bar{y})$, where $F(\bar{x},\bar{y})=G(\bar{x},\bar{y})=0$. Then, depending on the eigenvalues of the matrix of the linearized system
$$\begin{pmatrix}F_x & F_y\\G_x& G_x\end{pmatrix},\tag{2}$$
calculated at this point we conclude that it is either a node, or a saddle, or a focal point, or that linearization alone cannot tell us. Right?

In the integrable case we can construct function $H(x,y)$, s.t. it is constant along trajectories of (1), and therefore
$$F=\mu H_y, \qquad G=-\mu H_x \tag{3}$$
where $\mu=\mu(x,y)$ is and integrating factor and the stationary points of (1) are stationary (critical) of function $H(x,y)$. See Calculus II.  Remember that such points are of two types: maxima (or minima) and saddles. Again, some points where matrix of second derivatives (Hess matrix) cannot tell us the answer.

Assume  that this matrix is non-degenerate. Then either only maxima (minima) and level lines form center, or a saddle and level lines form a saddle.  Note matrix (2) at $(\bar{x},\bar{y})$ becomes
$$\mu \begin{pmatrix}H_{xy} & H_{yy}\\-H_{xx}& -H{x}\end{pmatrix},\tag{4}$$
with eigenvalues $\pm \mu \sqrt{H_{xy}^2-H_{xx}H_{yy}}$. Assume that $\mu(\bar{x},\bar{y})\ne 0$. Then

* If $H_{xy}^2-H_{xx}H_{yy}>0$ these eigenvalues are real and it is a saddle.

* If $H_{xy}^2-H_{xx}H_{yy}<0$ these eigenvalues are real and it is a center. In contrast to the general case linearization here gives us an answer (since no focal points are possible).  Only if $H_{xy}^2-H_{xx}H_{yy}=0$ at this point, we need to dig deeper... like in Calculus II.

One should remember this difference (very important). Now, check if the system in question is exact, if it is, find $H(x,y)$ and deal with it