Derivated function $f$ so that $f(x+y)=f(x)f(y)$ and $f'(x)f(y)=f(x)f'(y)$ for all $x,y \in \Bbb R$ Let $f:\Bbb R\to \Bbb R$ a derivated function in all $\Bbb R$ that satisfies the condition $$f(x+y)=f(x)f(y),\;\,\,\text{for all $x,y \in \Bbb R$}$$ 
I already tried that $f'(x)f(y)=f(x)f'(y)$ for all $x,y \in \Bbb R$ and that exists $c\in \Bbb R$ so that $f'(x)=cf(x)$ for all $x \in \Bbb R$ but from the above I conclude that tha function $f$ is $f(x)=e^{cx}$ but have not achieved this. I appreciate the help you can give me.

Let $x=y=0$ then $f(0)=f(0)^2$ if $f(0)=0$. Let $y=0$ we have $f(x)=f(x)f(0)=0$ 
Hence we have a function : $f \equiv 0$.
If $f(0)=1$ Let $y=-x$ we have $1=f(0)=f(x)f(-x)$.
I can't prove $f(x)>0$. If $f(x)>0$ then let $g(x)=\ln(f(x))$ and $g(0)=0$. Hence,
$g(x+y)=g(x)+g(y)$ thus, $\frac{g(x+y)-g(y)}{x}=\frac{g(x)}{x}=g'(c)$ with $c \in (x,x+y)$
hence $g(x)=g'(c)x=ax$
