real analysis - Differentiability of $f(x+y) = f(x)f(y)$

Let $f$: $\mathbb R$ $\to$ $\mathbb R$ be a function such that $f(x+y)$ = $f(x)f(y)$ for all $x,y$ $\in$ $\mathbb R$. Suppose that $f'(0)$ exists. Prove that $f$ is a differentiable function.

This is what I've tried: Using the definition of differentiability and taking arbitrary $x_0$ $\in$ $\mathbb R$.

$\lim_{h\to 0}$ ${f(x_0 + h)-f(x_0)\over h}$ $=$ $\cdots$ $=$ $f(x_0)$$\lim_{h\to 0}$ ${f(h) - 1\over h}$.

Then since $x_0$ arbitrary, using $f(x_0+0) = f(x_0) = f(x_0)f(0)$ for $y = 0$, can I finish the proof?