Find all continuous functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x, y \in \mathbb{R}$, we have:

$$
f(x+y)=f(x) f(y)
$$
Let $x=y$, then we have: $f(2 x)=f(x)^{2} \geq 0$. As $x$ runs through $\mathbb{R}$, $2 x$ also runs through $\mathbb{R}$. Therefore, $\forall x \in \mathbb{R}, f(x) \geq 0$.

If there exists $x_{0}$ such that $f\left(x_{0}\right)=0$, let $x=x_{0}$. Then, $f\left(x_{0}+y\right)=f\left(x_{0}\right) f(y)=0$ for all $y \in \mathbb{R}$. As $y$ runs through $\mathbb{R}$, $x_{0}+y$ also does. Therefore, $\forall y \in \mathbb{R}, f(y)=0$. We verify that the zero function is indeed a solution.

If for all $x \in \mathbb{R}, f(x) \neq 0$, then $\forall x \in \mathbb{R}, f(x)>0$. We can therefore compose both sides of the equality with the natural logarithm function. We then have:

$$
\ln (f(x+y))=\ln (f(x) f(y))=\ln (f(x))+\ln (f(y))
$$

Let $g=\ln \circ f$, then $g$ is a solution to Cauchy's equation. Since $g$ is the composition of two continuous functions, it is also continuous. Therefore, according to Cauchy's equation, $g(x)=c x$.

To find $f$, we compose both sides with the exponential function, yielding: $f(x)=\exp (c x)$. We verify this solution, which also works.