Trying to solve $f(x)f(y) = f(x + y)$ Let $f$ be a differentiable function $f: \mathbb{R} \mapsto \mathbb{R}$ such that 
$\forall x, y \in \mathbb{R}, \, f(x + y) = f(x)f(y)\tag{1}$
If $y=0$ then $f(x + 0) = f(x)f(0)$ which is just $f(x) = f(x)f(0)$. Then if $f(0) = 0$, we have $f(x) = f(x)f(0) = f(x) \cdot 0 = 0$ for all $x$. 
If $f(0) \neq 0$ instead then $f(0) = f(0 + 0) = f(0)f(0) = f(0)^2$ implies that $0 = f(0)^2 - f(0) = f(0)(f(0) - 1)$ which has solutions $f(0) = 0$ or $f(0) = 1$, but since we assume $f(0) \neq 0$, it must be the case that $f(0) = 1$. 
Furthermore let's assume the possibility of some real $z$ such that $f(z) = 0$. Then for all $x$ we have $f(x) = f(z + (x - z)) = f(z)f(x-z) = 0 \cdot f(x-z) = 0$. In other words, if the function equals $0$ at one point, it is zero for all of them. The contrapositive is that if the function is nonzero for any value, then it does not equal $0$ anywhere else. Therefore in the case where $f(0) = 1$, we have $f(x) \neq 0$ for all $x \in \mathbb{R}$.
Going forward we assume the case where
$f(0) = 1 \tag{2}$
By induction, $f(x)^n = f(nx)$ for all positive integers $n>0$ as follows. 
Base case with $n=1$, we have $f(x)^1 = f(1 \cdot x)$ which is $f(x) = f(x)$ which is trivially true.
Inductive step we assume $f(x)^n = f(nx)$ and we must show that $f(x)^{n+1} = f((n+1)x)$. 
$\begin{align}
f(x)^n &= f(nx) & \text{By inductive hypothesis}\\
f(x)^n f(x) &= f(nx)f(x) \\
f(x)^{n+1} &= f(nx + x) & \text{By equation (1)}\\
f(x)^{n+1} &= f((n+1)x)
\end{align}$
Thus for the positive integers we prove the result:
$\forall n \in \mathbb{Z^{+}}, \, f(x)^n = f(nx)\tag{3}$
Next we see that $f(x) = f(\frac{x}{2} + \frac{x}{2}) = f(\frac{x}{2})^2$ and since any real number squared is non-negative, we combine this with our assumption that $f(x) \neq 0$ for all $x \in \mathbb{R}$ and so it follows that 
$\forall x \in \mathbb{R}, \, f(x) > 0 \tag{4}$
so we are free to divide by $f(x)$ values safely without concerning ourselves with division by $0$ issues.
Since $1 = f(0) = f(x + (-x)) = f(x)f(-x)$ it follows that
$\forall x \in \mathbb{R}, \, f(x)^{-1} = f(-x) \tag{5}$
Then for some negative integer $n < 0$ we have $f(x)^n = (f(x)^{-n})^{-1} = f(-nx)^{-1} = f(nx)$ by $(3)$ and $(5)$. Therefore:
$\forall n \in \mathbb{Z^{-}}, \, f(x)^n = f(nx)\tag{6}$
We can also directly show that $f(x)^n = f(nx)$ holds for $n=0$, since $f(x)^0 = 1 = f(0 \cdot x) = f(0)$ which holds by assumption $(2)$. Now we can conclude:
$\forall n \in \mathbb{Z}, \, f(x)^n = f(nx)\tag{7}$
Now if we have $n \neq 0$ we can take the $n$th root of both sides of equation $(7)$ and get $f(x) = f(nx)^{1/n}$ which also implies that $f(x/n) = f(x)^{1/n}$. 
Then for integers $m, n$ with $n \neq 0$ we have $f(m \frac{x}{n}) = f(mx)^{1/n}$ using this result, and then applying equation $(3)$ we transform it to $(f(x)^m)^{1/n}$ or $f(x)^{m/n}$ (again for $n \neq 0$).
If we let some rational $r = m/n$ then we see that
$\forall r \in \mathbb{Q}, \forall x \in \mathbb{R}, f(rx) = f(x)^r \tag{8}$
By setting $x=1$ it is revealed that 
$\forall r \in \mathbb{Q}, \, f(r) = f(1)^r \tag{9}$
And then there's some theory I don't understand that I can apply here when $f$ is continuous (as assumed up front since we say $f$ is differentiable) that lets me jump to
$\forall x \in \mathbb{R}, \, f(x) = f(1)^x \tag{10}$
Is my proof right so far?
 A: There is no clarity on whether $x$ is fixed or variable when you say $f(x)\neq 0$. You can start by observing that if $f(x)=0$ for some $x$ then  $f(x+y)=0$ for all $y$ which implies $f(z)=0$ for all $z$. We we may assume that $f(x) \neq 0$ for all $x$. Except for this what you have done is correct. But you still have to complete the argument. Here is a hint: show that $f(rx)=(f(x))^{r}$ for any positive rational number $r$. Then use continuity to say that this equation holds for all positive real numbers $r$. Then put $r=\frac 1 x$. You will see that $f(x)=(f(1))^{x}$. So the answer is $f(x)=c^{x}$ for some $c$. 
A: Another approach might be to exploit the differentiability of $f(x)$ ...
$$f'(x) = \lim _{ h \to 0} \frac{ f(x+h)-f(x) }{ h }
\\=f(x)    \bigg (\lim _{ h \to 0} \frac{ f(h)-1 }{ h } \bigg )
\\=f(x)f'(0)$$
Let $k \equiv f'(0)$, so $f(x) $ is the solution to the differential equation ...
$$ f'(x)=kf(x) $$
satisfying the initial condition $f(0)=1$
The complete solution to this d.e. is 
$$ f(x)= e^{kx} $$
where $k$ can be any real number.
