Find $f(0)$ given some initial conditions.

Let's take a rigorous approach on this functional equation, shall we?

CASE I ( $f$ is constant): $C = C^2 \Rightarrow 0 = C^2 - C = C(C-1) \Rightarrow C = 0, 1$ . But we're given that $f(0) \neq 0$ , so $f(x) = 1$ is the only satisfactory constant function which in turn gives $f(0)=1$ .

CASE II ( $f$ is non-constant): With a leap of faith we'll assume $f$ is differentiable over all real $x$ . Differentiating the original functional equation with respect to $x$ and $y$ produces:

$f'(x+y) = f'(x)f(y)$ (i)

$f'(x+y) = f(x)f'(y)$ (ii)

which equating (i) with (ii) yields: $\frac{f'(x)}{f(x)} = \frac{f'(y)}{f(y)} = A \Rightarrow \ln f(x) = Ax+B \Rightarrow f(x) = e^{Ax+B}$ (iii). Substituting (iii) back into the functional equation now gives:

$e^{A(x+y)+B} = e^{Ax+B} \cdot e^{Ay+B};$

or $e^{A(x+y)+B} = e^{A(x+y)+2B} \Rightarrow B = 0.$

Hence $f(x) = e^{Ax}$ is the family of solutions that satisfies non-constant $f$ , which $f(0) = 1$ holds true.

Let $x=0$ .
$f(x+y)=f(x)f(y)$
$\Rightarrow f(0+y)=f(0)f(y)$
$\Rightarrow f(y)=f(0)f(y)$
$\Rightarrow f(0)=\frac{f(y)}{f(y)}$
$\Rightarrow f(0)=1$
So, $\boxed {f(0)=1}$

We set $x=y=0$ so we have:

$f(0+0)=f(0)f(0)\iff f(0)=\big(f(0)\big)^2\iff f(0)=\boxed1\;\text{ since }f(0)\neq0.$