Real Analysis Continuous Function Problem Show that the only continuous function on $(-1,+1)$, which is not identically
zero and satisfies the equation $f(x + y) = f(x)f(y)$ for all $x,y \in \mathbb{R}$, is the exponential function $f(x) = a^x$ with $a = f(1) > 0$.

This answer is basically in the same spirit as Hagen von Eitzen's answer but probably in a more "algorithmic" way. From his answer we know that $f(0)=1$.
Step 1: For all $n\in\mathbb N$, we have:
$$
f(2)=f(1)^2\\
\vdots\\
f(n)=f(1)^n
$$
So for all positive integers $n$ , we have $f(n)=a^n$ with $a=f(1)$.
Step 2: We know that $f(nx)=f(x)^n$ for all positive integers $n$. Therefore $f(\frac 1n)^n=a$. Also:
$$
f(\frac mn)=f(\frac 1n)^m=a^{\frac mn}
$$
So for all positive rationals $\frac mn$ , we have $f(\frac mn)=a^{\frac mn}$ with $a=f(1)$.
Step 3: For an arbitrary positive real number $r$, there is a sequence of $\{a_n\}$ of rational numbers such that $\lim_{n\to\infty}a_n=r$. Now $f$ is supposed to be continuous and we have:
$$
f(r)=f(\lim_{n\to\infty}a_n)=\lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}a^{a_n}=a^r.
$$
So for all positive real numbers $r$ , we have $f(r)=a^{r}$.
Step 4: Use $f(-x)=\frac 1{f(x)}$ to prove the result for all $r\in\mathbb R$. 

Usually, when we deal with a functional equation, we can use these steps to find the function. Frist we find the value of function for some obvious choice of inputs (like $x=0$ here). Then we try to find the function for all integers and then all rationals. Finally, if the function is assumed to be continuous, we can use a sequence of rationals to find the function at all irrational number. That's why I called this approach "algorithmic". No need to say that there is no universal solution applicable to all functional equations. 
