If $f$ is continuous and $f(x+y)=f(x)f(y)$, then $\lim\limits_{x \rightarrow 0} \frac{f(x)-f(0)}{x}$ exists

The following is a solution of the functional equation $$f(x + y) = f(x)f(y)$$ for all real $x, y$ in a more general setting where $f:\mathbb{R}\to \mathbb{R}$ is a function which is continuous everywhere.

One solution (the trivial one) is that $f(x) = 0$ for all $x$ and then the limit $\lim_{x \to 0}\dfrac{f(x) - f(0)}{x} = 0$ exists.

Let's us then suppose that $f(x)$ is not identically zero. So there is a number $a$ such that $f(a)\neq 0$. Now $f(a) = f(a/2)f(a/2) \geq 0$ so $f(a) > 0$. Now we can see that $$f(a) = f(a + 0) = f(a)f(0)$$ so that $f(0) = 1$.

Note further that if $f(b) = 0$ then $f(x) = f(b)f(x - b) = 0$ for all $x$. Hence it follows that $f(x) \neq 0$ for all $x$ and by the reasoning in last paragraph $f(x)$ is positive everywhere. Next let $f(1) = k$ where $k > 0$ and then I will show that the behavior of $f$ crucially depends on $k$. Note that by simple alegbraic arguments and using the equation $f(x)f(y) = f(x + y)$ we can show that $f(x) = k^{x}$ when $x$ is rational. And if $k = 1$ then $f(x) = 1$ for all rational $x$ and via continuity $f(x) = 1$ for all $x$. Thus if $k = 1$ the desired limit in question exists and is equal to $0$.

It is important to observe that the value of $f(1) = k$ determines the function uniquely. Thus if $g, h$ are two non-trivial continuous solutions of the functional equation $f(x + y) = f(x)f(y)$ then $g, h$ are positive and hence the function $F$ given by $F(x) = g(x)/h(x)$ is well defined and continuous. Moreover $F$ is also a solution of the functional equation. If $g(1) = h(1)$ then $F(1) = 1$ and by last paragraph $F(x) = 1$ for all $x$ and hence if $g(1) = h(1)$ then $g(x) = h(x)$ for all $x$. Thus we see that a continuous non-trivial solution of $f(x + y) = f(x)f(y)$ is uniquely determined by the value $k = f(1)$.

Next let $k > 1$. Using some amount of algebraic manipulation we can show that $$\frac{f(x) - f(0)}{x} < \frac{f(y) - f(0)}{y}$$ for rational $x, y$ with $0 < x < y$. If $x, y$ are real numbers with $0 < x < y$ then we can choose sequences of rationals $x_{n}, y_{n}$ such that $x_{n} \to x, y_{n} \to y$. Let $u, v$ be rationals with $0 < x < u < v < y$ and then after a certain value of $n$ we will have $0 < x_{n} < u < v < y_{n}$ and therefore $$\frac{f(x_{n}) - f(0)}{x_{n}} < \frac{f(u) - f(0)}{u} < \frac{f(v) - f(0)}{v} < \frac{f(y_{n}) - f(0)}{y_{n}}$$ and on taking limits as $n \to \infty$ we get $$\frac{f(x) - f(0)}{x} \leq \frac{f(u) - f(0)}{u} < \frac{f(v) - f(0)}{v} \leq \frac{f(y) - f(0)}{y}$$ Here we have used continuity of $f$. Note that from the above we get the desired equation $$\frac{f(x) - f(0)}{x} < \frac{f(y) - f(0)}{y}$$ for all real $x, y$ with $0 < x < y$ under the assumption that $f(1) = k > 1$. So we have established that the function $g(x) = \dfrac{f(x) - f(0)}{x}$ is strictly increasing in $(0, \infty)$.

We next need to show that $g(x) \geq 0$ for all $x > 0$. Again let $x_{n}$ be a sequence of rationals tending to $x$. Then $x_{n} > 0$ after a certain value of $n$ and then $$g(x_{n}) = \frac{f(x_{n}) - f(0)}{x_{n}} = \frac{k^{x_{n}} - 1}{x_{n}} > 0$$ because $k > 1$. Taking limits as $n \to \infty$ and via continuity of $g$ we get $g(x) \geq 0$ for all $x$. It follows now that $\lim_{x \to 0^{+}}g(x)$ exists.

We will show that $\lim_{x \to 0^{-}}g(x)$ also exists and is same as $\lim_{x \to 0^{+}}g(x)$. Clearly we have \begin{align} L &= \lim_{x \to 0^{-}}g(x)\notag\\ &= \lim_{x \to 0^{-}}\frac{f(x) - f(0)}{x}\notag\\ &= \lim_{t \to 0^{+}}\frac{f(-t) - f(0)}{-t}\notag\\ &= -\lim_{t \to 0^{+}}\frac{f(-t + 0) - f(-t + t)}{t}\notag\\ &= -\lim_{t \to 0^{+}}f(-t)\cdot\frac{f(0) - f(t)}{t}\notag\\ &= f(0)\lim_{t \to 0^{+}}\frac{f(t) - f(0)}{t}\notag\\ &= \lim_{x \to 0^{+}}\frac{f(x) - f(0)}{x}\notag \end{align} It follows that $\lim_{x \to 0}g(x)$ exists. If $0 < k < 1$ then also the same result holds but then $g(x)$ may be negative and you need to establish the bound that $g(x) \geq (k - 1)/k$ with more algebraic manipulation.

The value of the limit $$\lim_{x \to 0}\frac{f(x) - f(0)}{x}$$ crucially depends on $k = f(1)$ and thus defines a function of $k$ say $L(k)$. We have seen that for $k = 1$ we have $L(k) = 0$ and with some effort we can show that $L(x)$ is a continuous and differentiable function of $x$ on $(0, \infty)$ with the property that $L(xy) = L(x) + L(y)$ and $L'(x) = 1/x$ (you should try to prove these facts about $L$) and at this point you understand that $L(x)$ is more popularly known and written as $\log x$.

Note that in the above we have proved that $f'(0)$ exists and using functional equation $f(x)f(y) = f(x + y)$ we can show easily that $f'(x) = f(x)f'(0)$. Thus using the fact that $f$ is continuous and $f(x + y) = f(x)f(y)$ we have shown that $f$ is differentiable for all $x$. Going a step further (and also to summarize) we can show that

If $f$ is continuous at any single point say $a$ and $f(x + y) = f(x)f(y)$ then $f$ is continuous and differentiable everywhere with $f'(x) = f(x)f'(0)$ and leaving aside the trivial solution $f(x) = 0$ everywhere the value of $f'(0)$ is a function of $f(1) = b$ so that $f'(0) = L(b)$ and $L$ is continuous and differentiable on $(0, \infty)$ and $L'(x) = 1/x$.

You can easily check that $f(\frac pq) = f(1)^{\frac pq} \; \forall \frac pq \in \Bbb Q$, so $\left.f\right|_{\Bbb Q}$ is just an exponential.

Now, since we know the function is continous and $\Bbb Q$ happens to be dense in $\Bbb R$ we can state that $f(x) = f(1)^x \; \forall x$, and now you just have to compute the derivative of some $a^x$, which we know is differentiable at $x=0$.

To get around the fact that you can't differentiate $f$ (yet), we'll integrate it! Define $F$ to be the following primitive of $f$

$$F(x) = \int_0^x f(t) \, dt$$

Integrating $f(x+y) = f(x)f(y)$, with regards to $x$, between $0$ and $1$ (actually you may chose any non zero number here), we get

$$F(y+1) - F(y) = f(y)F(1)$$

Since $f$ is positive, $F(1) \ne 0$, and we get

$$f(y) = \frac{F(y+1)-F(y)}{F(1)}$$

Now the right hand side is clearly $\mathcal{C}^1$, so $f$ is $\mathcal{C}^1$ too.