Functions whose derivative is proportional to themselves

1. Let $g(x)=f(x)e^{-x}$. Then $g'(x)=f'(x)e^{-x}-f(x)e^{-x}=0$. A function with zero derivative on an interval is constant (mean value theorem), so $g=C$ and $f(x)=Ce^x$.

2. First, $f(0)=1$: choose $x_0$ with $f(x_0)\ne0$; then $f(x_0)=f(x_0+0)=f(x_0)f(0)$ gives $f(0)=1$.

Fix $y$ and differentiate $f(x+y)=f(x)f(y)$ in $x$:
$$f'(x+y)=f'(x)f(y).$$
Taking $x=0$: $f'(y)=f'(0)f(y)$. Let $a=f'(0)$. Then $f'=af$.

Apply part 1 to $x\mapsto f(x/a)$ if $a\ne0$, or directly: $g(x)=f(x)e^{-ax}$ has $g'=af e^{-ax}-af e^{-ax}=0$, so $g$ is constant, equal to $g(0)=1$. Hence $f(x)=e^{ax}$. (If $a=0$, $f'=0$ and $f=f(0)=1=e^{0\cdot x}$.)

Remark. The exponential is characterised by $f'=f$, $f(0)=1$, or by the functional equation: three faces of the same function.