Below this answer @WeiZhong commented that it is wrong, I also think it is wrong.
Reducing each numerator to $1$: $$f(x)=\cfrac{x}{x+\cfrac{x^2}{x^2+\cfrac{x^3}{x^3+\lower2ex\ddots}}}\\ =0+\cfrac{1}{1+\cfrac{x}{x^2+\cfrac{x^3}{x^3+\lower2ex\ddots}}}\\ =0+\cfrac{1}{1+\cfrac{1}{x+\cfrac{x^2}{x^3+\lower2ex\ddots}}}\\ =0+\cfrac{1}{1+\cfrac{1}{x+\cfrac{1}{x+\lower2ex\ddots}}}$$ Let's follow a Wikipedia article's notation, write the continued fraction as $$=[0;1,x,x,x^2,x^2,x^3,x^3,\dots]$$ $1+f(1)=[1;1,1,1,\dots]$ is well-known to be $\phi$.
Question:
Is $f(x)$ divergent on $0<x<1$?
Is $f(x)$ an analytic function on $x>1$?
I posted the question in SageMath community to plot $f(x)$. Plot for integers $x\in(1,10)$:
n_terms = 18
def f(x):
X = sum( ([x^i,x^i] for i in range(n_terms)), [])
return continuant([0]+X) / continuant(X)
plot(f, (1,10))
The question is about the continued fraction
$$ f(x)=x/(x+x^2/(x^2+x^3/(x^3+\dots))). \tag1 $$
My standard method is to make this precise. Define the continued fraction sequences by recursion
\begin{align*} f_n(x, q, x_0) &:= x / (x + f_{n\!-\!1}(x\,q, q, x_0)), &f_0(x, q, x_0) := x_0, \\ g_n(x, q, x_0) &:= x + 1/g_{n\!-\!1}(x\,q, q, x_0), &g_0(x, q, x_0) := x_0, \\ h_n(x, q, x_0) &:= 1 + x/(h_{n\!-\!1}(x\,q, q, x_0)), &h_0(x, q, x_0) := x_0. \tag2 \end{align*}
Verify the relations between the three sequences
$$ f_{2n}\Big(x, x, \frac{x^n}{x_0}\Big) = \frac1{g_n(1, x, x0)}, \;\; f_n\Big(\frac1x, \frac1x, \frac1{x_0}\Big) = \frac1{h_n(x, x, x_0)}. \tag3 $$
The infinite continued fraction in equation $(1)$ is the limit of $\,f_n(x, x, x_0)\,$ as $\,n\to\infty.\,$ For $\,0<x<1\,$ the sequence has two limits depending only on the parity of $\,n\,$ but for $\,1<x\,$ there is only one limit.
To be more precise, for $\,0<x<1,\,$ we get
$$ f_{2n}(x,x,1) = f_{2n+1}(x,x,0). \tag4 $$
Also,
$$ S(-x) := \lim_{n\to\infty} f_{2n}(x,x,1) = 1 - x + x^3 - x^5 + x^6 + x^7 - 2x^8 + 2x^{10} +\dots \tag5 $$
A search of the OEIS finds $\,S(x)\,$ is the generating function of OEIS sequence A003823. This implies
$$ S(x)=1+x/(1+x^2/(1+x^3/(1+\dots)))=f(-x^2,-x^3)/f(-x,-x^4) \tag6 $$
where $\,f(,)\,$ is the Ramanujan theta function.
For the case where $\,1<x, \,$ or equivalently, $\,0<\frac1x<1,\,$ the limit is
$$ R(x) := \lim_{n\to\infty} f_n\Big(\frac1x,\frac1x,1\Big) \!=\! 1 - x + x^2 - x^4 + x^5 - x^6 + x^7 - x^9 +2x^{10} + \dots. \tag7 $$
A search of the OEIS finds this is the generating function of OEIS sequence A007325 the Rogers-Ramanujan's continued fraction. This implies
$$ R(x) \!=\! \frac1{S(x)} \!=\! 1/(1+x/(1+x^2/(1+x^3\dots))) = f(-x, -x^4)/f(-x^2, -x^3). \tag8 $$