$ 0 = y - x^2, 0 = x - y^2$
$ x = x^4, (x,y) = (1, 1), (0, 0)$
$F = x - y^2, F_x = 1, F_y = -2y$
$G = y - x^2, G_x = -2x, G_y = 1$
at $(x, y) = (1, 1)$
$
\left[ {\begin{array}{cc}
1 & -2 \\
-2 & 1 \\
\end{array} } \right]
$
gives characteristic equation
$r^2 - 2r - 3 = 0, r = 3, -1$
Solutions are real but opposite sign so near (1, 1) it's a saddle point
at $(x, y) = (0, 0)$
$
\left[ {\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array} } \right]
$
$r = 1$
Solutions are real and positive. Near (0, 0) is a unstable node.