Prove that Cexp(x)C\exp(x) is the only set of functions for which f(x)=f′(x)f(x) = f'(x)

Of course $C e^x$ has the same property for any $C$ (including $C = 0$). But these are the only ones.

Proposition: Let $f : \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0) = 1$ and $f'(x) = f(x)$. Then it must be the case that $f = e^x$.

Proof. Let $g(x) = f(x) e^{-x}$. Then

$$g'(x) = -f(x) e^{-x} + f'(x) e^{-x} = (f'(x) - f(x)) e^{-x} = 0$$

by assumption, so $g$ is constant. But $g(0) = 1$, so $g(x) = 1$ identically.

N.B. Note that it is also true that $e^{x+c}$ has the same property for any $c$. Thus there exists a function $g(c)$ such that $e^{x+c} = g(c) e^x = e^c g(x)$, and setting $c = 0$, then $x = 0$, we conclude that $g(c) = e^c$, hence $e^{x+c} = e^x e^c$.

This observation generalizes to any differential equation with translation symmetry. Apply it to the differential equation $f''(x) + f(x) = 0$ and you get the angle addition formulas for sine and cosine.