Interesting functional equation: $f(x)=\frac{x}{x+f\left(\frac{x}{x+f(x)}\right)}$ Solve for the function f(x):

$$f(x)=\frac{x}{x+f\left(\frac{x}{x+f(x)}\right)}$$
I'm not able to solve this. 

[For instance, I tried solving for $f(\frac{x}{x+f(x)})$, but this doesn't lead me anywhere as the value obtained, when substituted into the original equation, just yields $f(x)=f(x)$]

Question: What is f(x) = ?

I suppose I need to mention that f(x) should be continuous, and that it's  a subset of $R$.

(Related : Functional equation $f(x)=\frac{1}{1+f(\frac{1}{1+f(x)})}$ , whose confirmed solutions are $-\phi$ and $\frac{1}{\phi}$)
If such an $f$ exists, plugging in $x=0$ we find that
$$f(0)=\frac{0}{0+f\left(\tfrac{0}{0+f(0)}\right)}=0,$$
and to avoid having to divide by $0$ this requires that
$$0+f(0)\neq0\qquad\text{ and }\qquad0+f\left(\frac{0}{0+f(0)}\right)\neq0.$$
But then $f(0)=0$ so $0+f(0)=0$, a contradiction. Hence no such $f$ exists.
take $g=\frac { x }{ (x+f) } $ (#)
by differentiating from two sides of this: $1=g+g'x+f'g+g'f$  $(*)$
by substituting $g=\frac { x }{ (x+f) } $ in the equation: $$\frac { x }{ g } -x=\frac { x }{ \left( x+f(g) \right)  } $$ 
differentiating from two sides of this and using (*) : $2x-gx=g-g^2$
find g in this quadratic equation.Then,find f by (#)