Prove $f'(x) = 2 f(x)$ if $f(x+y) = f(x) f(y)$, $f(x) \ne 0$ and $f'(0) = 2$. I'll state the question from my textbook below:

A function $f: \mathbb{R} \to \mathbb{R}$ satisfies the equation $f(x+y) = f(x) f(y)$ for all $x,y \in \mathbb{R}, f(x) \ne 0$. Suppose that the function is differentiable at $x = 0$ and $f'(0) = 2$. Prove that $f'(x) = 2f(x)$.

Firstly, I don't understand the first sentence completely. Does it mean that this equation holds true whenever $f(x) \ne 0$ or does it mean to say that $f(x) \ne 0, \forall x \in \mathbb{R}$. If it's the latter then $f(y) \ne 0$ too, right?
And then I proceeded in many different ways to find $f'(x)$. I tried replacing $y$ by $0$ in the given equation and then differentiated it, first differentiated the equation and then replaced $y$ by $0$ and tried a few other things. None of these got me anywhere.
Here's something I proved while trying to solve the question:
Putting $x,y = 0$ in the given equation we have:
$f(0) = f(0) f(0) \implies f(0)[f(0) - 1] = 0$
Since $f(x) \ne 0, \forall x \in \mathbb{R}$ (I'm not really sure if this is what the question meant, let's say it did),
$f(0) - 1 =0 \implies f(0) = 1$
I don't know if this is even useful but it seems to be so, since we also have been given $f'(0) = 2$.
Please help me prove the required equation. Any help would be appreciated.
 A: I'm sure it means $f(x)\ne0$ for all real $x$.
Now
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h=
\lim_{h\to0}\frac{f(x)f(h)-f(x)}h=f(x)\lim_{h\to0}\frac{f(h)-1}h$$
etc.
A: Regarding $x$ as fixed and differentiating $f(x+y)=f(x)f(y)$ with respect to $y$, we obtain
$$ f'(x+y) = f(x) f'(y). $$
Inserting $y=0$ and using $f'(0) = 2$ gives you the result.
A: Incidentally, the claim is still true if we interprete it as 

A function $f\colon\Bbb R\to \Bbb R$ satisfies the equation $f(x+y)=f(x)f(y)$ for all $x,y\in\Bbb R$ with $f(x)≠0$. Suppose that the function is differentiable at $x=0$ and $f′(0)=2$. Prove that $f′(x)=2f(x)$.

Proof: For $f'(0)$ to exist, $f$ must be continuous in a neighbourhood of $0$. If $f(0)\ne0$, then $f(x)\ne 0$ in a neighbourhood of $0$. And if $f(0)=0$, then $f'(0)=2$ implies $f(x)\ne 0$ in a punctures neighbourhood of $0$. So there certainly exists $r>0$ such that $f(x)\ne 0$ for all $x$ with $0<|x|<r$. But for such $x$, we obtain $f(0)=f(x+(-x))=f(x)f(-x)\ne0$, so also $f(0)\ne 0$. Assume that there exists $x$ with $f(x)=0$. Let $s=\inf\{\,|x|:f(x)=0\,\}$. Then $s\ge r>0$. For $s\le |x|<2s$, we find $f(x/2)\ne0$ and then $f(x)=f(x/2)^2\ne0$, contradicting the definition of $s$.
We conclude that $f(x)\ne 0$ for all $x$, hence in fact $f(x+y)=f(x)f(y)$ for all real $x,y$. 
With this, $$f'(y)=\frac{\mathrm d}{\mathrm dx}f(x+y)\mid_{x=0}=\frac{\mathrm d}{\mathrm dx}(f(x)f(y))|_{x=0}=f'(0)f(y)=2f(y).$$ 
