Function of single variable $f(x)$, $f(x+y)=f(xy)$ and the exponential. For a function of a single variable $f(x)$ which has the property that $f(x+y)=f(x)f(y)$ we first set $x=y=0$ and then develop an ODE to show that $f(x)=\exp\{-\beta x\}$. 
I do not understand how setting $x=y=0$ gives us $f(0)=1$ before we know that we have the exponential? Then I don't understand how we can set the following ODE up.
\begin{equation}
\frac{df(x)}{dx}=-\beta f(x)\ \ \ \ \ \ \ \text{where} \ \ \ \ \ \ \ \beta=-f'(x)
\end{equation}
Which doesn't contain any information on $y$. I do however understand that the solution to that ODE is the exponential I am after. Many thanks! 

For $\, x = y = 0\,$ we have $\, f(0+0) = f(0)\cdot f(0).\,$ 
Let us denote $\,f_0:=f(0),\,$ then
$$
f_0 = f_0^2 \implies 
\begin{cases}
f_0 = 1\\
f_0 = 0
\end{cases}
$$
Note that for any $x\in\Bbb R\,$ $\,f(x + 0) = f(x)\cdot f(0)$.
Thus we conclude that the second case $\,f_0 = 0\,$ is trivial. 
Indeed, if $\,f(0) = 0\,$ then 
$\,
f(x + 0) = f(x)\cdot f(0) = f(x)\cdot 0 = 0,\,
$
and thus $\,f(x) \equiv 0.\,$
Therefore we choose the first case $\,f_0 = f(0) = 1.\,$
