Solve $1-y = 0\Rightarrow y=1$ and $x^2 - y^2 = (x+y)(x-y) = 0$, substituting $y=1$, we have 2 critical points $(-1, 1), (1,1)$
Computing the Jacobian yields
\begin{align}
J =
\begin{bmatrix}
F_x & F_y\\
G_x & G_y
\end{bmatrix}
=
\begin{bmatrix}
0 & -1\\
2x & -2y
\end{bmatrix}
\end{align}
Plugging in both critical points we have 2 linear systems, first
\begin{align}
x' =
\begin{bmatrix}
0 & -1\\
-2 & -2
\end{bmatrix}
x
\end{align}
solving for eigenvalues yields $\lambda_1 = -1 + \sqrt{3}\, \lambda_2 = -1 - \sqrt{3}$, since $\sqrt{3} > 1$, we have $\lambda_1 > 0 > \lambda_2$, so we conclude locally $(-1, 1)$ is a saddle.
Plug in $(1,1)$
\begin{align}
x' =
\begin{bmatrix}
0 & -1\\
2 & -2
\end{bmatrix}
x
\end{align}
solving for eigenvalues yields $\lambda_1 = -1 + i\, \lambda_2 = -1 - i$, a complex conjugate with negative real parts, we conclude locally $(1, 1)$ is an asymptotically stable spiral.
