$f : \mathbb{R} \to \mathbb{R}$ $\forall_{x\in \mathbb{R}} f(x) * f(f(x))=1 (*) \\ f(1000)=999$ calculate $f(500)=?$
So $f(1000)*f(f(1000))=1 \to 999*f(999)=1 \to f(999)= \frac{1}{999}$ Darboux. $\exists x_0 \in (999,1000)$ With (*) $ x=x_0$ $f(x_0)*f(f(x_0))=1 \to $ $500*f(500)=1 \to $ $f(500)=\frac{1}{500}$ I don't know how to give an example of a function that meets the given conditions.
Any $y$ in the image of $f$ will have $y=f(x)$ for some $x \in \mathbb R$ and then satisfy $y \times f(y)=1$ so $f(y)=\frac1y$ and thus $f\left(\frac1y\right)=y$. Let's call $f$'s image $Y=\{y \in \mathbb R \mid \exists x \in \mathbb R: y=f(x)\}$
You now know that $0$ cannot be in $Y$. Since $f(1000)=999$, you also know $1000$ and $\frac1{1000}$ cannot be in $Y$ and that $999$ and $\frac1{999}$ must be in $Y$. Any such $f$ and $Y$ which meets these conditions with $f(y)=\frac1y$ for $y \in Y$ will be a solution. So $f(500)$ can potentially take any value apart from $0$ or $1000$ or $\frac1{1000}$ or $500$
To construct an example, suppose you want $f(500)=k \not \in \left\{0 , 1000,\frac1{1000},500 \right\}$. Then consider
$$ f(x) = \begin{cases} 999 &\quad\text{if } x= 1000 \text{ or } x= \tfrac1{999}\\ \tfrac1{999} &\quad\text{if } x= 999\\ \tfrac1{k} &\quad\text{if } x= k\\ k &\quad\text{otherwise.} \\ \end{cases}$$
There will be many other possible examples