Das relações do enunciado, temos:$$f(x)\cdot f(y) = \sin {(\log x)}\cdot \sin {(\log y)}$$e$$\frac{1}{2}\cdot[g\left(\dfrac{x}{y}\right)-g(x\cdot y)]~=~\dfrac{1}{2}\cdot[\cos \log \left(\frac{x}{y}\right)-\cos \log (x\cdot y)]$$esta última que, simplificada pelas regras de logaritmos, fica: $$\dfrac{1}{2}\cdot[\cos (\log x - \log y)-\cos (\log x + \log y)]$$Sabendo que$$\cos(a-b)-\cos(a+b)~=~2\sin a \sin b$$Então, simplificando mais ainda, obtém-se:$$\sin(\log x)\cdot \sin(\log y)$$Logo$$\frac{1}{2}\cdot[g\left(\frac{x}{y}\right)-g(x\cdot y)]~=~\sin(\log x)\cdot \sin(\log y)$$Então, concluí-se que:$$f(x)\cdot f(y) - \frac{1}{2}\cdot[g\left(\frac{x}{y}\right)-g(x\cdot y)]~=~\sin(\log x)\cdot \sin(\log y)-\sin(\log x)\cdot \sin(\log y)$$Portanto$$\boxed {f(x)\cdot f(y) - \frac{1}{2}\cdot[g\left (\frac{x}{y}\right)-g(x\cdot y)]~=~0}$$$$\bf{Alternativa~(E)}$$

$f(x) \cdot f(y) - \frac{1}{2}[g(\frac{x}{y}) - g(x \cdot y)]$ $= \sin(\log_{}x) \cdot \sin(\log_{}y) - \frac{1}{2}( \cos(\log_{}(\frac{x}{y})) - \cos(\log_{}(x \cdot y)) )$ $= \sin(\log_{}x) \cdot \sin(\log_{}y) - \frac{1}{2}( \cos(\log_{}x - \log_{}y) - \cos(\log_{}x + \log_{}y) )$ $= \sin(\log_{}x) \cdot \sin(\log_{}y) - \frac{1}{2}( \cos(\log_{}x) \cdot \cos(\log_{}y) + \sin(\log_{}x) \cdot \sin(\log_{}y) - \cos(\log_{}x) \cdot \cos(\log_{}y) + \sin(\log_{}x) \cdot \sin(\log_{}y))$ $= \sin(\log_{}x) \cdot \sin(\log_{}y) - \frac{1}{2}(2\sin(\log_{}x) \cdot \sin(\log_{}y) )$ $ = \sin(\log_{}x) \cdot \sin(\log_{}y) - \sin(\log_{}x) \cdot \sin(\log_{}y) = \boxed{0} $ $\textbf{Resposta : Alternativa E}$