Is there a function whose inverse is exactly the reciprocal of the function, that is $f^{-1} = \frac{1}{f}$? Is there a function whose inverse is exactly the reciprocal of the function?
That is $f^{-1} = \frac{1}{f}$.
We know that the inverse of a function is not necessarily equal to its reciprocal in general. 
 A: 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.

A: 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 $g$:
$$ g(x) = \begin{cases}
      \dfrac 1 2 + \dfrac 1 {n+1} & x = \dfrac 1  n, n \in \Bbb{N} \\
      \dfrac 1 2 + x & \text{otherwise} \\
\end{cases}
$$
A: 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.
A: For definiteness, let $X$ and $Y$ denote non-empty sets of complex numbers and let $f:X \to Y$ be a bijection whose inverse $f^{-1}:Y \to X$ is the reciprocal $1/f:X \to 1/Y$. (Here, "$1/Y$" denotes the set of reciprocals of elements of $Y$. Consequently, $X = Y = 1/Y$: The domain and target of $f$ are the same set, and are closed under taking reciprocals, and $0$ is not an element of $X$.)
At the level of elements, for all $x$ in $X$ and $y$ in $Y$,
$y = f(x)$ if and only if
$$
x = f^{-1}(y) = \frac{1}{f(y)}.
\tag{1}
$$
Substituting $y = f(x)$ in (1) and taking reciprocals,
$$
f^{[2]}(x) := f\bigl(f(x)\bigr)
  = \frac{1}{x}\quad\text{for all $x$ in $X$.}
\tag{2}
$$
Applying $f$ to both sides and using associativity of composition,
$$
f^{[3]}(x) = f\left(\frac{1}{x}\right) = \frac{1}{f(x)}\quad\text{for all $x$ in $X$.}
\tag{3}
$$
One more application of $f$ gives
$$
f^{[4]}(x) = f^{[2]}\left(\frac{1}{x}\right) = x\quad\text{for all $x$ in $X$.}
\tag{4}
$$
That is, $f$ generates an action of the cyclic group of order $4$ on $X$. Further, each point $x$ in $X$ "generates" a square in which the solid arrows denote application of $f$, and the dashed arrows represent application of the reciprocal function ($x \mapsto 1/x$).

Such functions certainly exist (modulo the axiom of choice, depending how large you want the domain $X$ to be); just pick distinct complex numbers $a$ and $b$ other than $0$, $1$, and $-1$, place $a$, $b$, $1/a$, and $1/b$ cyclically around the corners of such a square, and define $f$ according to the diagram. (You can also consistently define $f(1) = 1$ and/or $f(-1) = -1$, or $f(1) = -1$ and $f(-1) = 1$. These numbers are special because they're fixed by the reciprocal function itself.)

The question remains: Does any such function have an explicit algebraic formula? Offhand I don't know.
The function proposed in the comments, $f(x) = x^{i} = \exp(i\log x)$ with $\log$ denoting the principal branch, does not work without qualification: For $x > 0$ real,
$$
f\bigl(f(x)\bigr)
  = \exp\bigl(i\log(\exp i\log x)\bigr)
  \neq \exp(-\log x)
  = \frac{1}{x}
$$
unless $\log(\exp i\log x) = i\log x$, which happens if and only if $\exp(i\log x)$ has absolute argument less than $\pi$, if and only if $e^{-\pi} < x < e^{\pi}$.
The formula $f(x) = \exp(i\log x)$ does work on any reciprocal-invariant, $f$-invariant set. No real interval has this property, however, since $f$ is not real-valued on any real interval. Every open disk with $a > 0$ and $1/a$ lying on a diameter is reciprocal-invariant, but a bit of numerical experimentation suggests no such disk is $f$-invariant. (I haven't carefully checked.)
A: The complex function $f(x) = x^{\pm i}$ satisfies this property.
Its inverse is $f^{-1}(x) = x^{\mp i}$ since $f^{-1}\circ f(x) = \left(x^{\pm i}\right)^{\mp i} = x$.
One can see that $1/f(x) = x^{\mp i} = f^{-1}(x)$.
