From
$$y=\frac{1}{1+y} \tag{1}$$
the resulting quadratic equation should be
$$y^2+y−1=0 \tag{2}$$
with the two solutions being $y=\frac{−1\pm\sqrt5}{2}$.
It turns out that these values of $y$ are the fixed points of the function $f:\mathbb{R}\to\mathbb{R}$ where
$$f(x)=\frac{1}{1+x} \tag{3}$$
The continued fraction converges to whatever value the iterated function sequence
$$x,\ f(x),\ f(f(x)),\ f(f(f(x))), \ldots$$
converges to. The fixed points of $f$ are the only possible values to which the continued fraction can converge. Convergence of the continued fraction in the neighbourhood of either of the fixed points depends on whether the fixed point is attractive or not. Since $f$ is continuously differentiable in the neighbourhoods of both $\frac{−1-\sqrt5}{2}$ and $\frac{−1+\sqrt5}{2}$ (there is a discontinuity only at $-1$), either fixed point is attractive if and only if
$$|f'(x_0)|<1$$
From (3),
$$f'(x)=\frac{-1}{(1+x)^2} \tag{4}$$
Now at either fixed point, equation (1) is obeyed with $y=x_0$, so (4) reduces to
$$f'(x_0)=-\left(\frac{1}{1+x_0}\right)^2=-x_0^2 \tag{5}$$
So now calling $x_1=\frac{−1-\sqrt5}{2}$ and $x_2=\frac{−1+\sqrt5}{2}$ we must have
$$f'(x_1)=-\left(\frac{−1-\sqrt5}{2}\right)^2=\frac{-3-\sqrt5}{2} \implies |f'(x_1)|>1$$
so $x_1$ is an unstable fixed point, and the sequence will diverge away from this value.
Answer 1
Hence, the root $x_1=\frac{−1-\sqrt5}{2}$ arises as a non-attractive fixed point that can be safely disregarded as the continued fraction cannot possibly converge to this value.
For $x=\frac{-1-\sqrt5}{2}$ exactly, $f(x)=\frac{-1-\sqrt5}{2}$ but this is an unstable equilibrium (any perturbations cause a divergence).
Answer 2
Also,
$$f'(x_2)=-\left(\frac{−1+\sqrt5}{2}\right)^2=\frac{\sqrt5-3}{2} \implies |f'(x_2)|<1$$
so $x_2$ is an attractive fixed point. Because $f'(x_2)<0$, the sequence will converge in alternating fashion (oscillating about the fixed point) to the fixed point at $x_2=\frac{−1+\sqrt5}{2}$.
Since
$$\begin{align}
x\in(-\infty,-2) &\implies f(x)\in(-1,0) \\
x\in(-1,0) &\implies f(x)\in(0,\infty) \\
x\in(0,\infty) &\implies f(x)\in(0,\infty) \\
\end{align}$$
it is fairly straightforward to see that the sequence will converge to $\frac{-1+\sqrt5}{2}$ for $x\in(-\infty,-2)\cup(-1,0)\cup(0,\infty)$.
