My attempt: $f(x+y)=f(x)f(y)$
Differenting wrt $y$,
$$f'(x+y)=f(x)f'(y)$$
Putting $y=0$,
$$f'(x)=f(x)f'(0)=kf(x)$$
$$\frac{dy}{dx}=ky$$
We know, $y \neq 0$ because $y$ is different everywhere so dividing by y and integrating both sides
$$\int \frac{dy}{y}= \int dx$$
$$\ln \lvert y\rvert = kx + c$$
(We can get $c=0$ by putting $x=y=0$ in the first equation)
$$y = \pm {e}^{kx}$$
Doubt: Shouldn't we get $y = \pm {a}^{kx}$ as solution where $a$ $\in$ $R$ except $0$ ?
Actually, it does not matter as long as we are dealing with exponential functions.
You can still write the solution in terms of $a^{jx}$ where $a$ is a non-negative real number.
Take $j = \frac{k}{\log a}$ then you see that $e^k = a^j$, so they are interconvertible.
An alternative way. First find $f(0)$. Note that, for $y=0$ we have $$ f(x)=f(0)\cdot f(x), \mbox{ for all } x\in \mathbb{R} $$
If there is $x_{0}\in \mathbb{R}$ such that $f(x_{0})\neq 0$ then $$ f(0)=1. $$ Otherwise, $ f $ will be trivially indeed null.
We have $$f(x)\cdot f(-x)=1,\quad\mbox{ for all } x\in \mathbb{R}-\{0\}$$
If $n\in \mathbb{N}$ then $$f(n\cdot x) = f(x)^n,\quad \mbox{ for all } x\in \mathbb{R}-\{0\}$$
If $m\in \mathbb{Z}$ then $$f(m\cdot x) = f(x)^m,\quad \mbox{ for all } x\in \mathbb{R}-\{0\}$$
If $q\in \mathbb{Z}-\{0\} $ then $$f\left(\frac{1}{q}\cdot x\right) = f(x)^{\frac{1}{q}},\quad \mbox{ for all } x\in \mathbb{R}-\{0\}$$
If $q\in \mathbb{Z}-\{0\} $ and $p\in\mathbb{Z}$ then $$f\left(\frac{p}{q}\cdot x\right) = f(x)^{\frac{p}{q}},\quad \mbox{ for all } x\in \mathbb{R}-\{0\}$$
If $r\in\mathbb{Q}$ then $$f\left(r\right) = f(1)^{r}$$
We can choose $f(1)$ like any positive real number. As any $x\in \mathbb{R}$ is limit of a sequence of rational numbers $x=\lim_{n\to \infty}r_n$ and any exponential function is continuous $\lim_{n\to \infty}f(1)^{r_{n}}=f(1)^{\lim_{n\to \infty}r_{n}}$ we have $$ f(x)=f(1)^x $$
