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.

All you need is continuity at $0$.
If for some $c$, $f(c) = 0$, then it is easy to show that $f(x) = 0$ for all $x$.
So suppose $f(x) \neq 0$ for any $x$.
Now $f(x) = (f(x/2))^2 \gt 0$.
Thus it makes sense to talk about $g(x) = \log f(x)$.
This satisfies $g(x+y) = g(x) + g(y)$.
This is the Cauchy functional equation and continuity at $0$ implies $g(x) = cx$ and $f(x) = e^{cx}$
btw, to complete your work:
$f'(x) = cf(x) \implies f(x) = Ae^{cx}$
We basically get $ (f(x)e^{-cx})' = 0 \implies f(x)e^{-cx} = A \implies f(x) = Ae^{cx}$
Now you can show that $f(1) = 1$ (or $0$) and get either $f(x) = e^{cx}$ or $f(x) = 0$.
