This is exercise 6.26.8 from Tom Apostol's Calculus I, I'd like to ask someone to verify my proof. I'd be also interested in alternative proofs:
If $ f(x+y)=f(x)f(y) $ for all $ x $ and $ y $ and if $ f(x)=1+xg(x) $, where $ g(x) \to 1 $ as $ x \to 0 $, prove that (a) $f'(x)$ exists for every $x$, and (b) $f(x)=e^x$.
(a) $$ f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} = \lim_{h \to 0}\frac{f(x)f(h) - f(x)}{h} = \lim_{h \to 0}f(x)\frac{1 + hg(h) - 1}{h} = \lim_{h \to 0}f(x)g(h) = f(x) $$
(b) $$ \left(\frac{f(x)}{e^x}\right)' = \frac{f'(x)e^x - f(x)e^x}{e^{2x}} = \frac{f(x) - f(x)}{e^x} = 0 \implies f(x) = ke^x \; \text, \; k\in \mathbb R $$ $$ k = ke^0 = f(0) = \lim_{x \to 0}1 + xg(x) = 1 \implies f(x) = e^x $$
Yes, it's fine. You have less computations if you set $F(x)=f(x)e^{-x}$, so $$ F'(x)=f'(x)e^{-x}-f(x)e^{-x}=0 $$ and $F$ is constant.
An alternative proof could be by observing that $f(x)=0$ for some $x$ implies $f$ constant $0$, which contradicts the assumptions. So we know $f(x)\ne0$ for all $x$ and differentiability (part a) implies $f$ is continuous, so everywhere positive. Then $$ F(x)=\log f(x) $$ is well defined and $$ F'(x)=\frac{f'(x)}{f(x)}=1 $$