Functional equation: $f(x)f(1/x) = f(x) + f(1/x)$. If $x \neq 0$ , find $f(x)$ if it satisfies: $f(x)f(1/x)  = f(x) + f(1/x)$. 
I know that the answer is $f(x) = 1 \pm x^n$ where $n \in \mathbb{R}$. I don't know how to show this.
 A: Solution of $g(x)g(\frac 1 x)=1$ on $\mathbb R \setminus \{0\}$: take any function $h$ on $\{x: 0<|x|<1\}$ and define $g(x)=h(x)$ for $0<|x|<1$, $g(x)=\frac 1 {h(1/x)}$ for $|x| >1$ Take $g(1)$ and $g(-1)$ to be 1 or -1. Then $g$ satisfies the given property. Conversely, given $g$ we can take $h$ to be the restriction of $g$ to $0<|x|<1$. This describes all possible solutions.
A: Hint:
$$f(x)(f(1/x)-1)=f(1/x)$$
$$f(x)= \frac{f(1/x)}{f(1/x)-1}$$
$$f(x)-1 = \frac{1}{f(1/x)-1}$$
I.e. defining $g(x)=f(x)-1$, we have $g(x) = 1/g(1/x)$. Now solve this easier equation.
A: $f(x)f\left(\dfrac{1}{x}\right)=f(x)+f\left(\dfrac{1}{x}\right)$
$f(x)f\left(\dfrac{1}{x}\right)-f\left(\dfrac{1}{x}\right)=f(x)$
$(f(x)-1)f\left(\dfrac{1}{x}\right)=f(x)$
$f\left(\dfrac{1}{x}\right)=\dfrac{f(x)}{f(x)-1}$
$f\left(\dfrac{1}{x}\right)=\dfrac{1}{f(x)-1}+1$
$f\left(\dfrac{1}{x}\right)-1=\dfrac{1}{f(x)-1}$
Let $g(x)=f(x)-1$ ,
Then $g\left(\dfrac{1}{x}\right)=\dfrac{1}{g(x)}$
$g(x)g\left(\dfrac{1}{x}\right)=1$
In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe2111.pdf.
$\therefore g(x)=\pm e^{C\left(x,\frac{1}{x}\right)}$ , where $C(u,v)$ is any antisymmetric function.
$f(x)=1\pm e^{C\left(x,\frac{1}{x}\right)}$ , where $C(u,v)$ is any antisymmetric function.
