prove $f'(x)=f(x)$

Solution 1:

$$ \begin{align} f(x+h)-f(x) & = f(x)\cdot f(h) - f(x), \quad\text{from (a)} \\[8pt] & = f(x)(f(h)-1) \\[8pt] & = f(x)\cdot hg(h),\quad \text{from (b)} \end{align} $$

So $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}f(x)g(h)=f(x)$

Solution 2:

Let $x_0$ be an arbitrary real number.

Note that you have $f(0)=1$ from the second condition.

You have $$f'(x_0)=\lim\limits_{y\to 0} \frac{f(x_0+y)-f(x_0)}y = \lim\limits_{y\to 0} \frac{f(x_0)(f(y)-1)}y = f(x_0) \lim\limits_{y\to 0} \frac{f(y)-1}y = f(x_0)f'(0).$$

Since you wrote you have already proved this for $0$ and $1$, you know that $f'(0)=f(0)=1$. Thus the above equation is the same as $f'(x_0)=f(x_0)$.

Solution 3:

Assuming $f$ is differentiable:

You have that

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

Fix $x$ and differentiate wrt $y$.

$$f'(x+y)=f(x)f'(y)$$

Let $y=0$, and you get

$$f'(x)=f(x)f'(0)$$