If $f(x)=\cos(\log_ex)$, find the value of $(\frac{1}{x})f(\frac{1}{y})-\frac{1}{2}\{ f(xy)+f(\frac{x}{y}) \}$

Question

I did it like this : $$f\Bigg(\dfrac{1}{x}\Bigg) = \cos\Bigg(\log_e \Bigg( \dfrac{1}{x} \Bigg ) \Bigg)= \cos(\log_e1-\log_ex) $$$$ = \cos(0-\log_ex) = \cos(-\log_ex) = \cos(\log_ex)$$ Similarly, $$f\Bigg(\dfrac{1}{y}\Bigg) = \cos(\log_ey)$$ Now, $$f(xy) = \cos(\log_e(xy)) = \cos(\log_ex+\log_ey)$$ And $$f\Bigg( \dfrac{x}{y} \Bigg) = \cos\Bigg(\log_e\Bigg(\dfrac{x}{y}\Bigg)\Bigg) = \cos(\log_ex - \log_ey)$$ $$\therefore f(xy)f\Bigg(\dfrac{x}{y}\Bigg) = \cos(\log_ex+\log_ey)\cos(\log_ex-\log_ey) = $$$$ 2\cos(\log_ex)\cos(\log_ey)$$ $$\therefore \dfrac{1}{2} \Bigg (f(xy)+f\Bigg(\dfrac{x}{y}\Bigg) \Bigg ) = \cos(\log_ex)\cos(\log_ey)$$ $$\therefore f\Bigg(\dfrac{1}{x}\Bigg) f\Bigg(\dfrac{1}{y}\Bigg) - \dfrac{1}{2}\Bigg \{ f(xy) + f\Bigg(\dfrac{x}{y}\Bigg) \Bigg \} = $$$$ \cos(\log_ex)\cos(\log_ey)-\cos(\log_ex)\cos)(\log_ey) = 0$$

I think that I've done everything correctly but the options for the answer given in my textbook are as follows :

$\cos(x-y)$

$\log(\cos(x-y))$

$1$

$\cos(x+y)$

So, have I made a mistake somewhere or are the options in my book incorrect? I think the latter is true.

Thanks!