Where did the negative answer come from in the continued fraction $1+\frac{1}{1+1/(1+\dots)}$? In this question we increased solution domain by squaring both sides of equation but what about this one ? 
Here the question is to evaluate $1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots  }}}}}$
$$x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}}}$$
$$x-1=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots  }}}}}$$
$$\cfrac{1}{x-1}=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots }}}}}$$
$$\cfrac{1}{x-1}=x$$
$$x^2-x-1 = 0$$
$$x=\frac{1\pm \sqrt{5}}{2}$$
Now, it is obvious that answer can't be negative so : 
$$x=\frac{1+\sqrt{5}}{2}$$

Let $f(z)=1+\frac1z$. The function $f$ has two fixed points, which are $\frac{1\pm\sqrt{5}}{2}$.
At first, when you write
$$x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}}}$$
you're defining $x$ to be the particular fixed point of $f$ that attracts the orbit of $1$, if it exists. It so happens that $x$ does exist, and it equals $\frac{1+\sqrt{5}}{2}$.
Later, when you write
$$x=1+\frac1x$$
you're asserting that $x$ is some fixed point of $f$, but this equation doesn't specify which fixed point that is.
