Seems like we have here yet another vindication of Léo Sauvé's [famed dictum][2]...

I am going to sketch below the solution to this problem with which [A. C. Hindmarsh][1] came up and which was showcased on page 696 of the sixth issue of volume 76 (1969) of the *American Mathematical Monthly*. As far as I know, the problem was originally proposed by H. L. Nelson and appeared on page 779 of volume 75 (1968) of the *Monthly*: oddly enough, it would make its way to the problem & solutions column of the *Monthly* once again in 1976 (vol. 83, no. 4, p. 293); however, on that occasion it would be attributed to a Nathaniel Grossman of UCLA.

**Solution.** We wish to determine all functions $f \colon [0,\infty) \to [0,\infty)$ such that $f(0)=0$ and $f^{\prime}(x) = f^{-1}(x)$ for every $x \in I:=(0,\infty)$. We shall prove that, apart from the function mentioned by [Tsuyoshi Ito][3] in his reply, **there is no other function** $f$ **which satisfies all the constraints under consideration**.

First off, we notice that if $f$ is a function that *does the job*, then $f$ must be $\mathcal{C}^{1}$ and strictly increasing in $(0,\infty)$. Then, differentiating the identity $$f(f^{\prime}(x))=x$$
repeatedly, we obtain that $f$ is a function of class $\mathcal{C}^{\infty}$. What is more, we obtain that $f^{\prime \prime}>0$, $f^{\prime \prime \prime} < 0, \ldots, (-1)^{k}f^{(k)}>0$; it follows from [Bernstein's theorem on regularly monotonic functions][4] that $f$ **is a real-analytic function on** $(0,\infty)$.

Now, from the identity

$$\frac{d}{dx} f(f(x)) = f^{\prime}(f(x))f^{\prime}(x) = xf^{\prime}(x)$$

we get that

$$f(f(x)) = \int_{0}^{x} y \, f^{\prime}(y) \, dy$$ for every $x \in I$. This allows us to ascertain that $f$ has a fixed point $a \in I$: if this were not the case, the function $F \colon I \to \mathbb{R}$, defined for every $x \in I$ as $F(x)= f(x)-x$, would be of fixed sign. We claim that such a thing is not possible: indeed, if $f(x)>x$ for every $x \in I$, then $y = f^{\prime}(f(y)) > f^{\prime}(y)$ for every $y \in (0,x)$ and whence

$$ x < f(f(x)) = \int_{0}^{x} y \, f^{\prime}(y) \, dy < \int_{0}^{x} y^{2} \, dy = \frac{x^{3}}{3},$$

which doesn't necessarily hold when $x$ is sufficiently small; since the assumption that the inequality $f(x)<x$ holds for every $x \in I$ allows us to derive a similar contradiction, we conclude that any solution $f$ to the functional-differential in question does have a fixed point $a\in I$. Further, the strict convexity of $F$ implies that $F$ has at most two zeros, counting the one it has at $x=0$. Thus, $f$ **has exactly one fixed point** $a\in I$, with $f(x)<x$ in $(0,a)$, $f(x)>x$ in $(a,\infty)$, $f^{\prime}(x)>x$ in $(0,a)$, and $f^{\prime}(x)<x$ in $(a,\infty)$.
 
The uniqueness of the solution to the problem is established resorting to the fixed point whose existence has just been proven. Let us suppose that $f_{1}$ and $f_{2}$ are two functions satisfying all the constraints under consideration and let $g:=f_{1}-f_{2}$. Moreover, let us denote with $a_{i}$ the unique fixed point of $f_{i}$ in the interval $I$. Without loss of generality, we can suppose that $a_{1} \geq a_{2}$. The possibility that $a_{1}>a_{2}$ leads us to a contradiction (I am not adding all the details behind the corresponding analysis for the moment. I consider this is the least transparent part of Mr. Hindmarsh's argument. ). If $a_{1}=a_{2}=a$, then it is not difficult to convince oneself that $0=g(a)=g^{\prime}(a) = g^{\prime \prime}(a)  =\ldots$; being $g$ a real-analytic function in $I$, the latter equalities implies that $g$ vanishes identically and we are done. **QED.**

  [1]: https://www.genealogy.math.ndsu.nodak.edu/id.php?id=44518
  [2]: http://mathoverflow.net/a/145824/1593
  [3]: http://mathoverflow.net/a/34061/1593
  [4]: http://www.ams.org/journals/proc/1975-047-02/S0002-9939-1975-0354974-3/S0002-9939-1975-0354974-3.pdf