Simple Substitution, Right?

Relevant wiki: Logarithms

Substituting the function in the given expression we get; $\begin{aligned} \mathfrak E&=\cos(\log x)\cos(\log y)-\dfrac 12\left(\cos(\log \dfrac xy)+\cos(\log xy)\right)\\ &=\cos(\log x)\cos(\log y)-\dfrac 12\left(\cos(\log x- \log y)+\cos(\log x +\log y)\right)\\ \end{aligned}$

Substitute $\log x = A$ and $\log y = B$ , we get;

$\begin{aligned} \mathfrak E&=\cos A\cos B-\dfrac 12(\cos A \cos B- \sin A \sin B+\cos A \cos B + \sin A\sin B)\\ &=\cos A\cos B-\dfrac 12\left(2\cos A\cos B\right)\\ &=\boxed 0 \end{aligned}$