How to solve the functional equation $f(2x)=2f(x)/(1+(f(x))^2)$ 
solve the functional equation $f(2x)=\dfrac{2f(x)}{1+(f(x))^2}$

I've already known that $\tanh(2x)=2\tanh(x)/(1+(\tanh(x))^2)$.
hyperbolic tangent
 A: Hint:   let $\displaystyle g(x) = \frac{1+f(x)}{1-f(x)}\,$, then:
$$\require{cancel}
g(2x)=\frac{1+f(2x)}{1-f(2x)} = \frac{1+ \frac{2f(x)}{1+f^2(x)}}{1- \frac{2f(x)}{1+f^2(x)}}=\frac{(1+f(x))^2}{(1-f(x))^2} = g^2(x)
$$
With suitable smoothness assumptions, the solutions will be of the form $\,g(x)=e^{\lambda x}\,$.

[ EDIT ]  To elaborate a bit on the choice of $\,g\,$, which was neither arbitrary, nor blind luck... It goes back to the observation that $\,\tanh(x)\,$ satisfies the given functional equation, as the OP duly noted.

But $\displaystyle\, \tanh(x)= \frac{e^{2x}-1}{e^{2x}+1} \iff e^{2x} = \frac{1+\tanh(x)}{1-\tanh(x)}\,$, so the function $\displaystyle\, \frac{1+f(x)}{1-f(x)}\,$ might, with any luck, behave "like" an exponential and yield an easier-to-solve functional equation. Which it did.
A: To use the identity you have observed i.e. $\tanh(2x)=2\tanh(x)/(1+(\tanh(x))^2)$
Let $ g(x)= \tanh^{-1} \circ f(x)$, so
\begin{align*}
\dfrac{2f(x)}{1+(f(x))^2}&= \dfrac{2\tanh\circ g (x)}{1+(\tanh\circ g (x))^2}                   \\
&=    \tanh(2g(x))   
\end{align*}
and
$$f(2x)=\tanh\circ g (2x) $$
Therefore,
$$2g(x)=g(2x).     \fbox{1} $$
Any $\mathbb{Q}-$linear function satisfies the previous. To show that $g$ is $\mathbb{R}-$ linear assume that $g$ is differentiable, and $g'$ is continuous at zero.
Differentiating $\fbox{1}$, we obtain $g'(x)=g'(2x)$. Inductively, we obtain 
$g'(x)= g'(x\cdot 2^{-n})\rightarrow g'(0)$. Therefore,
$$g(x)=g'(0)x.$$
