Let find $y$ such that $x+y=xy\Rightarrow y=\frac{x}{x-1}$.
One can take $y=\frac{x}{x-1}$ for $x\neq 0$, $x\neq 1$, then
$$f\left(x+\frac{x}{x-1}\right)=g\left(\frac{1}{x}+\frac{x-1}{x}\right)\cdot\left(x\cdot \frac{x}{x-1}\right)^{2008}\Rightarrow$$
$$f\left(\frac{x^2}{x-1}\right)=g(1)\cdot \left(\frac{x^2}{x-1}\right)^{2008}$$
$t=\frac{x^2}{x-1}$ could be any in $(-\infty;0)\cup[4;+\infty)$, so at least for such $t$: $f(t)=g(1)\cdot t^{2008}$.
Let find $y$ such that $x+y=-xy\Rightarrow y=-\frac{x}{x+1}$.
One can take $y=-\frac{x}{x+1}$ for $x\neq 0$, $x\neq -1$, then
$$f\left(x-\frac{x}{x+1}\right)=g\left(\frac{1}{x}-\frac{x+1}{x}\right)\cdot\left(-x\cdot \frac{x}{x+1}\right)^{2008}\Rightarrow$$
$$f\left(\frac{x^2}{x+1}\right)=g(-1)\cdot \left(\frac{x^2}{x+1}\right)^{2008}$$
$t=\frac{x^2}{x+1}$ could be any in $(-\infty;-4]\cup(0;+\infty)$, so at least for such $t$: $f(t)=g(-1)\cdot t^{2008}$.
For any $t>4$: $f(t)=g(-1)\cdot t^{2008}=g(1)\cdot t^{2008}$, therefore $g(-1)=g(1)=a$.
For any $t\neq 0$: $f(t)=a t^{2008}$.
Taking $y=-x$ one can obtain $f(0)=g(0)=0$, so $f(t)=a t^{2008}$ for any $t$.
Taking $y=1/x$: $f(x^2+1/x)=g(x^2+1/x)$
Equation $x^2+1/x=t$ has solution for $x$ for any $t\in(-\infty;-2]\cup[2;\infty)$, so $f(t)=g(t)$ for any $t$ such that $|t|\geq 2$.
Taking $y=x$: $f(2x)=g(2/x)\cdot x^{4016}$.
Using $x=t$ and $x=1/t$ in last equation, one can get $f(2t)=g(2/t)\cdot t^{4016}$ and $f(2/t)=g(2t)\cdot (\frac{1}{t})^{4016}$. Multiplying these equations gives $f(2t)\cdot f(2/t)=g(2/t)\cdot g(2t)$.
Using $t$ such that $|2t|\geq 2$ last equation transforms to $f(2t)\cdot f(2/t)=g(2/t)\cdot f(2t)$. $a\neq 0$, then $f(2/t)=g(2/t)$. If $|2t|\geq 2$ then $|t| \geq 1$, $|1/t| \leq 1$, $|2/t| \leq 2$. So $f(x)=g(x)$ for any $x\neq 0$ such that $|x|\leq 2$.
So $g(x)=f(x)$ for any $x$ except case $a=0$.
If $a=0$ then $f(x)=0$ and $g(x)=0$ because $t=\dfrac{1}{x}+\dfrac{1}{y}$ can take any value $t\neq 0$ and $g(0)=0$.
The final answer: $f(x)=g(x)=a x^{2008}$, $a$ is arbitrary real constant.
