Problem: Find all functions $f \colon \RR \to \RR$ such that for all $x,y \in \RR$,
\[ f(\left\lfloor x\right\rfloor y) = f(x)\left\lfloor f(y)\right\rfloor. \]

Solution: The only solutions are $f(x) \equiv c$,
where $c = 0$ or $1 \le c < 2$.
It's easy to see these work.

Plug in $x=0$ to get $f(0) = f(0) \left\lfloor f(y) \right\rfloor$,
so either
\[ 1 \le f(y) < 2 \quad \forall y
  \qquad\text{or}\qquad f(0) = 0 \]

In the first situation,
plug in $y=0$ to get $f(x) \left\lfloor f(0) \right\rfloor = f(0)$,
thus $f$ is constant.
Thus assume henceforth $f(0) = 0$.

Now set $x=y=1$ to get
\[ f(1) = f(1) \left\lfloor f(1) \right\rfloor \]
so either $f(1) = 0$ or $1 \le f(1) < 2$.
We split into cases:
\begin{itemize}
  \ii If $f(1) = 0$, pick $x=1$ to get $f(y) \equiv 0$.
  \ii If $1 \le f(1) < 2$,
  then $y=1$ gives
  \[ f(\left\lfloor x \right\rfloor) = f(x) \]
  from $y=1$, in particular $f(x) = 0$ for $0 \le x < 1$.
  Choose $(x,y) = \left( 2, \half \right)$ to get
  $f(1) = f(2) \left\lfloor f\left( \half \right) \right\rfloor = 0$.
\end{itemize}