Sejam $156^x = a$ e $156^y = b$, assim: $f(x+y) = \large{\frac{ab\space +\space \frac{1}{ab} }{2}}$ e $f(x-y) = \large{\frac{\frac{a}{b}\space +\space \frac{b}{a} }{2}}$ implica:$$f(x+y)+f(x-y) = \color{green}{\frac{ab\space +\space \frac{a}{b} + \frac{b}{a} + \frac{1}{ab}}{2}}$$$$2f(x)f(y) = 2\cdot \frac{a\space +\space \frac{1}{a} }{2}\cdot \frac{b\space +\space \frac{1}{b} }{2} = \color{green}{\frac{ab\space +\space \frac{a}{b} + \frac{b}{a} + \frac{1}{ab}}{2}}$$ Demonstra-se, portanto, que $\boxed{f(x+y) + f(x-y) = 2f(x)f(y)}$ $\mathbb{C.Q.D.}$