Find all continuous functions (did I do this correctly?) 
Find all the continuous functions $f:\mathbb{R}\to\mathbb{R}$ satisfying:
$f(x+y)=f(x)+f(y)+f(x)f(y)$ for all $x,y\in\mathbb{R}$

Solution attempt:
$$\begin{align*}
f(x+y) + 1 &= f(x) + f(y) + f(x)f(y) + 1\\
f(x+y) + 1 &= (f(x) + 1)(f(y) + 1)
\end{align*}$$
$g(x) = f(x) + 1$, and $g(y) = f(y) + 1$
$$\begin{align*}
g(x+y) &= g(x)g(y)\\
(x,y) &= (x,x)\\
g(2x) &= g^2(x)\\
g(x) &= g^2(x/2)\\
g(x) &= c^x,\;\text{ where }c = e^a
\end{align*}$$
So $f(x) = c^x - 1$ for all x
 A: Your result is correct, however, there is a gap in proving that $g(x+y)=g(x)g(y)$ implies that $g(x)=c^x$.
First of all, $g(x)=g^2(\frac{x}{2})$ implies that $g(0)=0$ , or $g(0)=1$.


*

*If $g(0)=0$ then $f(0)=-1$ and using the given relationship for $y=0, x\in \mathbb R$, we get:


$$f(x+0)=f(x)+f(0)+f(x)f(0)\Rightarrow f(x)=f(x)-1-f(x)\Rightarrow f(x)=-1,\forall x\in \mathbb R$$


*

*Now, If $g(0)=1$ , we have to follow some steps in order to reach the
desired conclusion:

*

*For $n\in\mathbb N$ it is easy to see that $$g(n)=g(1+\dots+1)=g(1)\times\dots\times g(1)=[g(1)]^n$$

*Also note that $$g(1+(-1))=g(1)g(-1)\Rightarrow g(0)=g(1)g(-1)\Rightarrow g(-1)=\dfrac{1}{g(1)}=[g(1)]^{-1}$$

*Now for a negative integer $-m$ we have :
$$g(-m)=g(-1\dots-1)=g(-1)\times\dots\times g(-1)=[g(1)]^{-m}$$

*Next, prove it for $\dfrac{1}{n},n\in \mathbb N$. You have already proved that $g(2x)=g^2(x).$ If you extend this, you can easily get $g(nx)=g^n(x)$. So for $x=\frac{1}{n}$
$$g^n(\frac{1}{n})=g(1)\Rightarrow g(\frac{1}{n})=[g(1)]^{1/n}$$

*Now the big step: Prove it for rationals $\dfrac{m}{n}$
$$g(\frac{m}{n})=g(\frac{1}{n}+\dots+\frac{1}{n})=g(\frac{1}{n})\times\dots\times g(\frac{1}{n})=g(\frac{1}{n})^m=[[g(1)]^{1/n}]^m=[g(1)]^{m/n}$$

*Finally, if you take a  $x\in\mathbb R$, then there exists a sequence of rationals $\{q_n\}_{n=1}^\infty$ such that $q_n\xrightarrow{n\rightarrow \infty}x$ and:
$$g(x)=\lim\limits_{n\rightarrow\infty}g(q_n)=\lim\limits_{n\rightarrow\infty}[g(1)]^{q_n}=[g(1)]^x$$
where we used the continuity of $g$ as well as the continuity of the exponential function.



So we reached the desired result : $$g(x)=[g(1)]^x=c^x$$
$$\Rightarrow f(x)=g(x)-1=c^x-1$$
(or $f(x)=-1$ do not forget!)
A: Essentially yes, except that it's not immediately clear, as xpaul notes, that $g(x)=c^x$ follows from the previous line, especially since you didn't mention continuity. (Given you have $c=e^a$ but no mention of $a$, I suspect you left something out!)

Personally I proceed from $g(x+y)=g(x)g(y)$ by noticing $g(x)=g(x/2)^2\ge 0$ so either $g(x) = 0$ (and $f(x) = -1$ - note that if this happens at one point, it happens everywhere; why?) or
$$h(x)=\ln g(x) \quad \implies \quad h(x+y) = h(x) + h(y) \quad \implies \quad h(nx) = nh(x)$$
What next? Hint: Given any $x\in\mathbb R$ notice that we can approximate $x\approx \frac{m}{n}$ by a rational number. Answer below.

By making this approximation with some accuracy, $x-\frac{m}{n} = \delta$, we find
$$h(x) = h\left(\frac{m}{n}\right) + h(\delta) = m h\left(\frac{1}{n}\right) + h(\delta) = \frac{m}{n} h(1) + h(\delta) = x h(1) + h(\delta) - \delta h(1)$$
This is all exact. Use the continuity of $h$, making $\delta$ nice and small, and you find $h(x) = x h(1)$ which gives you the result you need. I'll leave you to fill in the details.

You can definitely do this in lots of ways; this is just the one which comes most naturally to my mind.
