I've tried many different things myself, but I can't find any exact function that seems to follow this path.
First, I tried thinking that $f(1)$ would most likely also be 1, so then $f(1)f(1/2)=1$, which narrows it down to a function where $f(1/2)=1$.
After this, I was thinking it might be parabolic, because if $f(1)=f(1/2)$ then the vertex would be along the line $x=\dfrac{3}{4}$, which leads to an entire infinite set of functions that allow it.
This is where I started feeling confused, because I wasn't sure how to find the coefficients of such a parabolic function, or how to prove it was a solution to all real inputs.
Edit: solution below
if f(1)f(1/2)=1, then both f(1) and f(1/2) must be 1, but f(1/2) must exist so that f(1/2)f(0)=1/2, but f(0) has to equal zero.
Evaluating the given identity at $x=\frac12$ and $x=-\frac12$ yields $f(\frac12)f(0) = \frac12$ and $f(-\frac12)f(-1) = -\frac12$; multiplying these together yield $$ \bigl( f(0)f(-\tfrac12) \bigr) f(0)f(-1) = -\tfrac14 \ne 0. $$ However, the given identity at $x=0$ gives $f(0)f(-\frac12) = 0$, which contradicts the above calculation. Therefore there are no such functions.