Suppose $x$ is a root of $p(z) = z^2 - b z - c$. Then, diving $p(z)$ over $z$ and solving that for $z$ gets us $$ z = b+ \frac{c}{z} $$ Iterating: $$ z = b + \cfrac{c}{b + \cfrac{c}{z}} = \cfrac{b}{c + \cfrac{c}{b+ \ddots}} $$ Since $\frac{1-\sqrt{5}}{2}$ is a root of $z^2 - z -1$ we have: $$ \frac{1-\sqrt{5}}{2} = -\frac{1}{\frac{1+\sqrt{5}}{2}} = - \cfrac{1}{1 + \cfrac{1}{1+ \frac{1}{1+\ddots}}} $$
Added: Consider a sequence, defined by
$x_{n+1} = b + \frac{c}{x_n}$, with
$x_0 = \frac{c}{y}$. Few initial terms of the sequence are
$\frac{c}{y}$,
$b + y$,
$b + \cfrac{c}{b+y}$,
$b + \cfrac{c}{b + \cfrac{c}{b+y}}$, etc. It is well known that terms of this sequence can also be obtained as a ratio of two solutions,
$a_n$ and
$b_n$ to the following recurrence equation: $$ v_{n} = b v_{n-1} + c v_{n-2} \tag{1} $$ with initial conditions
$a_0=y$,
$a_1 = c$ and
$b_0 = 1-\frac{b}{c} y$,
$b_1 = y$. Then
$x_n = \cfrac{a_n}{b_n}$. The solution to
$(1)$ has the form: $$ v_{n} = v_0 \frac{z_1 z_2^n - z_2 z_1^n}{z_1-z_2} + v_1 \frac{z_1^n - z_2^n}{z_1-z_2} $$ where
$z_1$ and
$z_2$ are the two roots of
$z^2 - b z -c = 0$. Assume
$z_2>z_1$. In the large
$n$ limit, $$ \lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{a_n}{b_n}= \lim_{n\to \infty} \frac{a_1 (z_1^n - z_2^n) + a_0 (z_1 z_2^n - z_1^n z_2)}{b_1 (z_1^n - z_2^n) + b_0 (z_1 z_2^n - z_1^n z_2)} = \frac{a_0 z_1 - a_1}{b_0 z_1 - b_1} = \frac{c (c- y z_1)}{c (y - z_1) + b y z_1} $$ In the case at hand
$x_0 = \infty$ and
$x_1 = b$, corresponding to the limit of
$y \to \infty$, in which case the value of the continued fraction becomes: $$ \lim_{n \to \infty} x_n = \lim_{y \to 0} \frac{c (c- y z_1)}{c (y - z_1) + b y z_1} = - \frac{c}{z_1} = z_2 $$