Showing that $f(x)=\frac x{x+1}$ is the unique function satisfying $f(x)+f\left(\frac1x\right)=1$ and $f(2x)=2f\big(f(x)\big)$ We are given a function $ f : \mathbb Q ^ + \to \mathbb Q ^ + $ such that $$ f ( x ) + f \left( \frac 1 x \right) = 1 $$
and
$$ f ( 2 x ) = 2 f \big( f ( x ) \big) \text . $$
Find, with proof, an explicit expression for $f(x)$ for all positive rational numbers $x$.
Every number I have evaluated is of the form $ f ( x ) = \frac x { x + 1 } $ and this clearly fits the functional equations, but I can't prove that it's the only solution. Can anyone help me? I have put down the start of my workings which led me to the conjecture of $ f ( x ) = \frac x { x + 1 } $.
Plugging in $ x = 1 $ clearly gives $ f ( 1 ) = \frac 1 2 $ and $ f ( 2 ) = 2 f \big( f ( 1 ) \big) = 2 f \left( \frac 1 2 \right) $ which we can plug back into the first equation to get that $ f ( 2 ) = \frac 2 3 $. Working in this vein I have been able to show that $ f ( x ) = \frac x { x + 1 } $ for particular values of $ x $, but not in general.
The most difficult part appears to be proving it for the even integers. To prove $ x = 8 $, we have
$$ f ( 12 ) = 2 f \left( \frac 6 7 \right) = 4 f \left( \frac 3 { 10 } \right) = 4 - 4 f \left( \frac { 10 } 3 \right) \\
= 4 - 8 f \left( \frac 5 8 \right) = 8 f \left( \frac 8 5 \right) - 4 = 16 f \left( \frac 4 9 \right) - 4 \\
= 32 f \left( \frac 2 { 11 } \right) - 4 = 64 f \left( \frac 1 { 12 } \right) - 4 = 60 - 64 f ( 12 ) \text , $$
giving us $ f ( 12 ) = \frac { 12 } { 13 } $. This will probably be the main area of difficulty in the proof.

A bit too long for a comment, but here's a start:
You have shown $f(1) = 1/2$, $f(2) = 2/3$ and hence $f(1/2) = 1/3$.
Plug $x = 1/2$ into your second identity to get $f(1) = 2f(f(1/2))$, i.e $1/2 = 2 f(1/3)$, i.e $\color{blue}{f(1/3) = 1/4}$. Thus, $\color{blue}{f(3) = 3/4}$ as well.
To get $f(4)$ and $f(1/4)$, we must use the second identity twice as well as the first identity. First $f(4) = 2f(f(2)) = 2f(2/3)$. In addition, $f(2/3) = 2f(f(1/3)) = 2f(1/4)$. Thus $f(4) = 4f(1/4)$ and since $f(4) + f(1/4) = 1$ by the first identity, $\color{blue}{f(4) = 4/5}$ and $\color{blue}{f(1/4) = 1/5}$. We also get $\color{blue}{f(2/3) = 2/5}$ and $\color{blue}{f(3/2) = 3/5}$.
Then $f(1/2) = 2f(f(1/4))$ gives $1/3 = 2f(1/5)$ to get $\color{blue}{f(1/5) = 1/6}$ and $\color{blue}{f(5) = 5/6}$.
To get $f(6)$ and $f(1/6)$, we try to repeat the arguments used for $f(4)$ and $f(1/4)$. First, $f(6) = 2f(f(3)) = 2f(3/4)$. In addition $f(2/5) = 2f(f(1/5)) = 2f(1/6)$. Alas, this is where my ad hoc calculations break down.
Hopefully, these simple calculations are useful to help others find the underlying structure necessary to continue.
