If $f\left(x\right)=\cos \left({\log }_{e}x\right),$ then $f\left(x\right)f\left(y\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f\left(xy\right)\right]$ has value 

$\begin{array}{l}f\left(x\right)=\cos \left(\log x\right)\\ \mathrm{or} f\left(x\right)f\left(y\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f\left(xy\right)\right]\\ =\cos \left(\log x\right)\cos \left(\log y\right)-\frac{1}{2}\left[\cos \left(\log x-\log y\right)\right]+\cos \left(\log x+\log y\right)]\\ =\cos \left(\log x\right)\cos \left(\log y\right)-\frac{1}{2}\left[2\cos \left(\log x\right)\cos \left(\log y\right)\right]\\ =0\end{array}$