= Solution
Assume $X,Y$ are continuously differentiable near the equilibrium. Put $\xi=x-a$, $\eta=y-b$ and expand the <vector field>:
$$
\frac d{dt}\begin{pmatrix}\xi\\\eta\end{pmatrix}=M\begin{pmatrix}\xi\\\eta\end{pmatrix}+o\!\left(\sqrt{\xi^2+\eta^2}\right),\qquad
\boxed{M=\begin{pmatrix}X_x(a,b)&X_y(a,b)\\Y_x(a,b)&Y_y(a,b)\end{pmatrix}.}
$$
This <Jacobian matrix> defines the linearized system. For distinct real <eigenvalues>, two negative values give a stable node, two positive values an unstable node, and opposite signs a <saddle point of a dynamical system>. A complex-conjugate pair with negative real part gives a stable focus, and one with positive real part an unstable focus. Thus a hyperbolic saddle has \b[two real <eigenvalues> of opposite signs], equivalently $\det M<0$.
The <linearization stability theorem> transfers these stability conclusions to the nonlinear system when no <eigenvalue> has zero real part. Distinctness alone is insufficient: a zero eigenvalue or a purely imaginary conjugate pair leaves a <nonhyperbolic equilibrium>, for which the nonlinear terms must be examined. In the purely imaginary case the linearized trajectories form a centre, but that does not establish a nonlinear centre.
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