Find a possible f(x) and g(x) Do you think that it is possible to find a $f$ and a $g$ such that $f(x) \neq x$ and  $\forall x \gt 1, y \gt 1$  then
$$f\left({1\over 1-{1\over x}-{1\over y}}\right) = {1\over 1-{1\over g(x)}-{1\over g(y)}}$$
EDIT 1:
and 
$$f(2x) = 2f(x)$$
$$g(2x) = 2g(x)$$
EDIT 2:
and 
$$f(x) \gt 0$$
$$g(x) \gt 0$$
 A: (began answering before the $f(2x)=2f(x),g(2x)=2g(x)$ was added)
$$f\left({1\over 1-{1\over x}-{1\over y}}\right) = {1\over 1-{1\over g(x)}-{1\over g(y)}}$$
Let $A(x)=\frac{1}{f\left(\frac{1}{x}\right)}, B(x)=\frac{1}{g\left(\frac{1}{x}\right)},u=\frac{1}{x},v=\frac{1}{y}$. Then:
$$\frac{1}{A(1-u-v)}=\frac{1}{1-B(u)-B(v)}\implies A(1-u-v)=1-B(u)-B(v)$$
Now, let $C(x)=1-A(1-x)$, then:
$$C(u+v)=B(u)+B(v)$$
Letting $u+v=x$, this implies that $\forall \alpha, B(u)+B(x-u)$ must be constant in $u \implies \frac{d}{du}(B(u)+B(x-u))=0$
Thus $B'(u)-B'(x-u)=0$ for all $u$ and all $x$, meaning that $B'$ is constant and thus that $B(x)=mx+b$ for some $m,b$. This would lead to $C(x)=mx+2b$, and, working backwards:
$$\implies A(x)=1-m(1-x)-2b=mx+(1-2b-m)$$
$$\implies f(x)=\frac{x}{m+(1-2b-m)x}, g(x)=\frac{x}{m+bx}$$
[I realise I've made assumptions about differentiability - currently thinking about how to remove them]
A: From $f(2x)=2f(x)$ and $g(2x)=2g(x)$, we can deduce that $\forall n\in\mathbb Z : f(2^nx)=2^nf(x)$ and $g(2^nx)=2^ng(x)$. 
So from
$$f\left({1\over 1-{1\over x}-{1\over y}}\right) = {1\over 1-{1\over g(x)}-{1\over g(y)}}$$
when $x=y$, and using $f(2x)=2f(x)$ and $g(2x)=2g(x)$, we can get to
$$f\left(\frac{1}{x-1}+1\right)=\frac{1}{g(x)-1}+1\quad (1)$$
In fact, using our first deduction, we obtain a more general form
$$f\left(\frac{1}{x-{1\over2^n}}+2^n\right)=\frac{1}{g(x)-{1\over2^n}}+2^n,\quad \forall n\in\mathbb Z\quad (2)$$
Since both $f$ and $g$ are continuous, we can work with limits. In $(2)$, when $n\rightarrow\infty$, the result is also infinite. More generally, this means that $f(x)\rightarrow\infty$ as $x$ approaches infinity.
Knowing this, let's try $x\rightarrow1^+$ in $(1)$ and solve for $g(1)$. We get $g(1)=1$. Thus $g(2^n)=2^n\quad \forall n\in\mathbb Z$.
My input doesn't really answer your question, but I hope it gives some insight for what to expect.
