functional equation

f(x) =0 is a trivial solution

f(x) = x/(1+ax) is the non-trivial solution

We have

f \left(\frac{1}{1+x} \right) = \frac{f\left(\frac{1}{x} \right)}{1+f\left(\frac{1}{x} \right)}

\Rightarrow \frac{1}{f \left(\frac{1}{1+x} \right)} = \frac {1}{f\left(\frac{1}{x} \right)}+1

So if g(x) = \frac {1}{f\left(\frac{1}{x} \right)}

we have g(x+1) = g(x)+1 \Leftrightarrow g(x) = x+k

Hence

f\left(\frac{1}{x} \right) = \frac{1}{ x+k}

and

f(x)= \frac{x}{ 1+kx}