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

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.