We shall show that only the identity function and the constant functions satisfy the conditions of the problem. It is clear that these functions verify indeed the conditions.

Because $f$ is continuous, its range $\{f(x) \mid x \in \mathbb{R}\}$ is an interval $I \subseteq \mathbb{R}$. If $I$ is degenerate at a point then $f$ is constant.

If $I$ is non-degenerate, let $a = \inf I < \sup I = b$, where $a, b \in \bar{\mathbb{R}}$. By the given condition we deduce that the restriction of $f$ to the interval $(a, b)$ is the identity:
$$f(x) = x, \quad a < x < b. \tag{1}$$
We shall show that $a = -\infty$ and $b = +\infty$, i.e. $I = \mathbb{R}$ and $f$ is the identity function. Suppose $a$ is a finite number. By the continuity of $f$ in $a$ and by (1) we get $f(a) = a$, so
$$f'(a) = f'_d(a) = \lim_{x \to a,\ x > a} \frac{f(x) - f(a)}{x - a} = \lim_{x \to a,\ x > a} \frac{x - a}{x - a} = 1. \tag{2}$$
On the other side $f$ has a minimum at $a$, because
$$f(a) = a = \inf I = \inf\{f(x) \mid x \in \mathbb{R}\},$$
so, by the Theorem of Fermat, $f'(a) = 0$, in contradiction with (2). We conclude $a = -\infty$. Analogously $b = +\infty$.

Therefore, the only differentiable functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f \circ f = f$ are the constant functions and the identity function.