Are there any non-constant differentiable functions $f : \mathbb R \to \mathbb R$ where each $t \in \mathbb R$ has $f(t)f(f'(t))=1$? Partition 1st differentiable $f : \mathbb R \to \mathbb R$, by the number of $t \in \mathbb R$ where $f(t)f(f'(t))=1$.
My main question is

*

*Are there any non-constant functions where every $t \in \mathbb R$ satisfies the property?

A few things I've noticed:

*

*I was able to generate all of the countable equivalence classes using $\text {exp, sin,}+,\circ,\cdot$.

*The property is volatile with respect to adding a constant $c$ to a given function $f$.

*I'm 99% sure $f$ cannot both be a finite-degree polynomial and have the property of the main question.

 A: As Eero has shown, if $t\in\mathbb R$ the solutions $f(t)$ can only take the constant values $\pm1$. However, if we restrict the domain to not include $0$ we can find non-constant solutions. Starting from the equation
\begin{align}\tag{1}\label{1}
f(f'(t))=\frac{1}{f(t)},
\end{align}
substitute $f'(t)$ for $t$ to see that
\begin{align}
f(f'(f'(t)))&=\frac{1}{f(f'(t))}=f(t),
\end{align}
so then $f'(t)$ is an involution, i.e. $f'(f'(t))=t$.
Taking the derivative of the equation \ref{1} we get that
\begin{align}
f'(f'(t))f''(t)&=\frac{-f'(t)}{f(t)^2}\quad\longrightarrow\quad
tf''(t)=\frac{-f'(t)}{f(t)^2},
\end{align}
an easily solvable ODE. Integrating we arrive at
\begin{align}\tag{2}\label{2}
tf'(t)-f(t)&=\frac{1}{f(t)}+C_1.
\end{align}
Separating, integrating, and rearranging we arrive at the 'general solution' for the three cases depending on the value of $C_1$:
\begin{align}\tag{3}\label{3}
(f^2+C_1f+1)\left(\frac{2f+\sqrt{C_1^2-4}+C_1}{2f-\sqrt{C_1^2-4}+C_1}\right)^{C_1/\sqrt{C_1^2-4}}=C_0t^2;\quad[C_1^2>4],\\\\
(f^2+C_1f+1)\exp\left(\frac{-2C_1}{\sqrt{4-C_1^2}}\arctan\left(\frac{2f+C_1}{\sqrt{4-C_1^2}}\right)\right)=C_0t^2;\quad [C_1^2<4],\\\\
(f+1)^2\exp\left(\frac{+2}{f+1}\right)=C_0t^2;\quad [C_1=2].
\end{align}
Note that if we take $t=0$, each solution degenerates to $f(t)\equiv1$ or $f(t)\equiv-1$. For brevity, I'll denote the general solution as $\Omega(C_1,f(t))=C_0t^2$ here.
Given some initial time $t_i$, the constants $C_1$ and $C_0$ are given by the system
\begin{align}
\Omega(C_1,f(t_i))=C_0t_i^2,\quad t_if'(t_i)=f(t_i)+\frac{1}{f(t_i)}+C_1,\quad f(t_i)f(f'(t_i))=1.
\end{align}
Which can be reduced to the two equations
\begin{align}\tag{4}\label{4}
t_i^2\sqrt{\frac{\Omega(C_1,1/f(t_i))}{\Omega(C_1,f(t_i))}}=f(t_i)+\frac{1}{f(t_i)}+C_1,\quad C_0=\frac{1}{t_i^2}\Omega(C_1,f(t_i)).
\end{align}
As a sanity check, here's a simple example. If our initial condition states that $f(t_i)=1$ we see this corresponds with
\begin{align}
C_1^2>4\ \ \text{if}\ \ t_i^2>4,\quad C_1^2=4 \ \ \text{if} \ \ t_i^2=4,\quad \text{or}\ \ C_1^2<4\ \ \text{if}\ \ t_i^2<4.
\end{align}
So given $f(\sqrt2)=1$, we see that $C_1=0$, $C_0=1$, and from the second case in equation \ref{3} our solution comes out to $f(t)^2+1=t^2$. This function satisfies equation \ref{1} and the initial conditions, but is not defined (within the reals) for $t^2<1$.
This problem had all the fun stuff!
Edit
I was working on another problem and had some inspiration for a somewhat neater parametric solution.
If we resume from the equation \ref{2}, we can transform to a linear, constant-coefficient second-order ODE
\begin{align}
tff'-f^2-C_1f-1=0,\\
zff'_z+f^2+C_1f+1=0;\quad [t=1/z],\\
z''_{\xi \xi}+C_1z'_{\xi}+z=0;\quad [f=z'_{\xi}/z].
\end{align}
The solution to which is
\begin{align}
z&=A_1e^{r_1 \xi}+A_2e^{r_2\xi};\quad \left[C_1^2\neq4\right],\\
z&=(A_1+A_2\xi)e^{\pm\xi};\quad \left[C_1^2=4\right],
\end{align}
where
\begin{align}
r_{1}=\frac{1}{2}\left(-C_1+\sqrt{C_1^2-4}\right)\quad \text{and}\quad r_{2}=\frac{1}{2}\left(-C_1-\sqrt{C_1^2-4}\right).
\end{align}
Then the parametric solution is easily found as
\begin{align}\tag{4} 
f(\xi)=\frac{A_1r_1e^{r_1\xi}+A_2r_2e^{r_2\xi}}{A_1 e^{r_1\xi}+A_2e^{r_2\xi}},\quad t(\xi)=\frac{1}{A_1 e^{r_1\xi}+A_2e^{r_2\xi}},\quad[C_1^2\neq4]\\
\\
f(\xi)=\frac{A_2}{A_1+A_2\xi}\pm1,\quad t(\xi)=\frac{1}{(A_1+A_2\xi)e^{\pm\xi}},\quad[C_1^2=4].
\end{align}
The parametric solution has three constants, which can be determined from equations \ref{1}, \ref{2}, and \ref{4}.
A: The property that every $t\in\mathbb{R}$ satisfies
\begin{equation}\tag{1}\label{functional-identity}
f(t)f(f'(t))=1
\end{equation}
is very restrictive. In particular, we can deduce uniqueness of extremal values to show that the maximum and minimum values of $f$ are equal, so $f$ must be constant.
Observation 1: Every critical point $t\in\mathbb{R}$ has the same value $f(t)=1/f(0)$.
Observation 2: If $f'(t)\neq 0$ for all $t\in(a,b)$, then $f$ is strictly monotone on the interval $(a,b)$.
Proof: By the intermediate value property, we deduce that $f'$ has constant sign on $(a,b)$. Hence $f$ is either strictly increasing or strictly decreasing on $(a,b)$, depending on the sign of $f'$. $\square$
Observation 3: If $f'(x)=0$ for some $x\in\mathbb{R}$, then on each of the half-open intervals $(-\infty,x]$ and $[x,\infty)$, $f$ is either constant or strictly monotone.
Proof: Suppose $f$ is non-constant on $[x,\infty)$. If $f$ is not strictly monotone on $[x,\infty)$, then by continuity we can find a smaller interval $(a,b)\subset [x,\infty)$ where $f$ is also not strictly monotone and $f(t)\neq f(x)$ for all $t\in(a,b)$. By Observation 2 we deduce that $f'(t)=0$ for some $t\in(a,b)$. But then we have a critical point with $f(t)\neq f(x)=1/f(0)$, contradicting Observation 1. $\square$
Observation 4: If $f$ is strictly monotone on an interval $(a,b)$, then $f'$ is also strictly monotone on $(a,b)$.
Proof: On an interval being strictly monotone is equivalent to being injective. If $f'(x)=f'(y)$, then $f(x)=1/f(f'(x))=1/f(f'(y))=f(y)$. Hence the injectivity of $f$ on $(a,b)$ implies injectivity of $f'$ on $(a,b)$. $\square$
By Observations 2-4 we can fix some $x\in\mathbb{R}$ such that both $f$ and $f'$ are monotone on both of the intervals $(-\infty,x]$ and $[x,\infty)$. This monotonicity allows us to constrain limiting behavior as $t\to\pm\infty$.
Observation 5: If $\lim\limits_{t\to\infty}f(t)=c$ exists for some finite $c\in\mathbb{R}$, then $c=1/f(0)$. Similarly $\lim\limits_{t\to-\infty}f(t)=1/f(0)$ if the limit exists and is finite.
Proof: Since $f'$ is monotone on $[x,\infty)$, if the limit $\lim\limits_{t\to\infty}f(t)$ is finite then $\lim\limits_{t\to\infty}f'(t)=0$. Therefore
$$\lim\limits_{t\to\infty}f(t) = \lim\limits_{t\to\infty} 1/f(f'(t)) = 1/f(0).$$
The claim for $t\to-\infty$ is identical using monotonicity on $(-\infty,x]$. $\square$
Observation 6: $f(t)$ has a finite limit as $t\to\pm\infty$.
Proof: By monotonicity, the limits always exist. We claim that they cannot be infinite. For instance, if $f(t)\to\pm\infty$ as $t\to\pm\infty$, then the identity \eqref{functional-identity} forces $f(f'(t))\to 0$. However by Observation 5, $f$ cannot have a finite limit equal to 0, giving a contradiction. $\square$
By Observations 5 and 6, we see that $\lim\limits_{t\to\pm\infty}f(t)=1/f(0)$. Combining this with Observation 1, we find that both the minimum and maximum values that $f$ attains are all equal to the same constant $1/f(0)$. Hence $f$ must be one of the two constant functions $f\equiv 1$ or $f\equiv -1$.
