Solutions of $f(x)\cdot f(y)=f(x\cdot y)$ Can anyone give me a classification of the real functions of one variable such that $f(x)f(y)=f(xy)$? I have searched the web, but I haven't found any text that discusses my question. Answers and/or references to answers would be appreciated.

There is a classification of the functions $f:\mathbb R\to\mathbb R$ satisfying
$$
f(x+y)=f(x)+f(y), \quad\text{for all $x,y\in\mathbb R$}. \qquad (\star)
$$
These are the linear transformations of the linear space $\mathbb R$ over the field $\mathbb Q$ to itself. They are fully determined once known on a Hamel basis of this linear space (i.e., the linear space $\mathbb R$ over the field $\mathbb Q$). 
This in turn provides a classification of all the functions $g:\mathbb R^+\to\mathbb R^+$ satisfying 
$$
g(xy)=g(x)g(y), \quad\text{for all $x,y\in\mathbb R^+$},
$$
as they have to be form $g(x)=\mathrm{e}^{f(\log x)}$, where $f$ satisfies $(\star)$. Note that $g(1)=1$, for all such $g$.
Next, we can achieve characterization of functions $g:\mathbb R\to\mathbb R^+$ satisfying 
$$
g(xy)=g(x)g(y), \quad\text{for all $x,y\in\mathbb R$},
$$
as $g(-x)=g(-1)g(x)$, which means that the values of $g$ at the negative numbers are determined once $g(-1)$ is known, and as $g(-1)g(-1)=g(1)=1$, it has to be $g(-1)=1$. Also, it is not hard to see that only acceptable value of $g(0)$ is $0$. 
Finally, if we are looking for $g:\mathbb R\to\mathbb R$, we observe that, if $g\not\equiv 0$, and $x>0$, then $g(x)=g(\sqrt{x})g(\sqrt{x})>0$. Thus $g$ is fully determined once we specify whether $g(-1)$ is equal to $1$ or $-1$.
Note that if $g: \mathbb R\to\mathbb R$ is continuous, then either $g\equiv 0$ or $g(x)=|x|^r$ or $g(x)=|x|^r\mathrm{sgn}\, x $, for some $r>0$. 
