Is there a function whose inverse is exactly the reciprocal of the function, that is $f^{-1} = \frac{1}{f}$?
Assuming that the function maps $\Bbb{R} \to \Bbb{R}$ and is differentiable, the answer is no.
Here's why. In order for a continuous function to be invertible, it must be strictly monotonic (monotone increasing or monotone decreasing). In fact:
CLAIM 1: If $f$ is an increasing function, then $f^{-1}$ is too.
PROOF: This is a straightforward, standard calculation using the chain rule and the fact that $f(f^{-1}(x)) = x$ for all $x$: $$ \frac{d}{dx} f(f^{-1}(x)) = f'(f^{-1}(x)) \cdot (f^{-1})'(x) \quad\text{and}\quad \frac{d}{dx} x = 1 $$ so $$ (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}. $$ According to this last expression, the signs of $(f^{-1})'$ and $f'$ are the same.
CLAIM 2: If $f$ is an increasing function, then $(1/f)(x)$ is decreasing.
PROOF: Calculate the derivative using the chain rule: $$ \frac{d}{dx} \frac{1}{f(x)} = -\frac{f'(x)}{(f(x))^2}. $$ Since the denominator is positive, the signs of $(1/f)'$ and $f'$ must be opposite.
It is possible for this to be satisfied in $\Bbb{R} \to \Bbb{R}$ everywhere except $ 0$.
$$ f(x) = \begin{cases} -x & x \geq 0 \\ -{x}^{-1} & x \lt 0 \\ \end{cases} $$
EDIT: User Slade asked whether it is possible in $\mathbb{R}^+ \to\mathbb{R}^+$. I've constructed a solution.
Let $ g(x) $ be a bijective function from $ \left(0,\frac{1}{2}\right] $ to $ \left(\frac{1}{2},1\right) $.
Then the following function satisfies $ f^{-1} = \dfrac{1}{f} $:
$$ f(x) = \begin{cases} g(x) & 0 \lt x \leq \frac{1}{2} \\ \dfrac{1} {g^{-1}(x)} & \frac{1}{2} \lt x \lt 1 \\ 1 & x = 1 \\ g^{-1}(\frac{1}{x}) & 1 \lt x \lt 2 \\ \dfrac{1} {g\left(\frac{1}{x}\right)} & 2 \leq x \\ \end{cases} $$
EDIT 2: Here's an example of such a $g$:
$$ g(x) = \begin{cases} \dfrac 1 2 + \dfrac 1 {n+1} & x = \dfrac 1 n, n \in \Bbb{N} \\ \dfrac 1 2 + x & \text{otherwise} \\ \end{cases} $$
This is by no means a complete answer, but it is a bit long for a comment.
If $f\circ 1/f = 1/f \circ f = x$, we have the equations:
$$f(1/f(x)) = x$$
$$f(f(x)) = 1/x$$
In particular, $f$ is a half-iterate of $1/x$. Note that there cannot possibly be a solution defined at $x=0$.
One observation is that if $f$ satisfies the above equations, $1/f$ does as well. So if we could somehow prove that there was at most one solution to the above equations, then we could conclude that $f=1/f$ and arrive at a contradiction.
This problem deserves a very clear specification about what domain $f$ should be defined on, whether it should be continuous, and so forth. For example, the answer $x^i$ works as a function on the complex plane, but it can only be a continuous function away from a branch cut. And the much more interesting question of whether there exists such a function $\mathbb{R}^+ \to\mathbb{R}^+$ is still open.