Find all monotonically increasing or monotonically decreasing functions $f: \mathbb{R}_+\to\mathbb{R}_+$ which satisfy the equation $f\left(xy\right)\cdot f\left(\frac{f\left(y\right)}{x}\right)=1$ for any two numbers $x$ and $y$ from $\mathbb{R}_+$.

Hereby, $\mathbb{R}_+$ is the set of all positive real numbers.

[i]Note.[/i] A function $f: \mathbb{R}_+\to\mathbb{R}_+$ is called [i]monotonically increasing[/i] if for any two positive numbers $x$ and $y$ such that $x\geq y$, we have $f\left(x\right)\geq f\left(y\right)$.
A function $f: \mathbb{R}_+\to\mathbb{R}_+$ is called [i]monotonically decreasing[/i] if for any two positive numbers $x$ and $y$ such that $x\geq y$, we have $f\left(x\right)\leq f\left(y\right)$.
**
   For any \( x \in \mathbb{R}_+ \), pick \( y \) such that \( M > xy > \frac{1}{M} \):
   \[
   M \geq \frac{f(x)}{y} \geq \frac{1}{M}
   \]
   This implies:
   \[
   f(x) = \frac{1}{x}
   \]
   is a working solution.

10. **Verification:**
    - For \( f(x) = 1 \):
      \[
      f(xy) \cdot f\left(\frac{f(y)}{x}\right) = 1 \cdot 1 = 1
      \]
    - For \( f(x) = \frac{1}{x} \):
      \[
      f(xy) = \frac{1}{xy}, \quad f\left(\frac{f(y)}{x}\right) = f\left(\frac{1/y}{x}\right) = f\left(\frac{1}{xy}\right) = xy
      \]
      \[
      f(xy) \cdot f\left(\frac{f(y)}{x}\right) = \frac{1}{xy} \cdot xy = 1
      \]

Thus, the solutions are \( f(x) = 1 \) and \( f(x) = \frac{1}{x} \).

The final answer is \( \boxed{ f(x) = 1 } \) or \( f(x) = \frac{1}{x} \).