The function given is \( f(x) = \frac{a^x + a^{-x}}{2} \). We compute \( f(x+y) \) and \( f(x-y) \) as follows:
\[ f(x+y) = \frac{a^{x+y} + a^{-(x+y)}}{2}, \quad f(x-y) = \frac{a^{x-y} + a^{-(x-y)}}{2}. \]
Adding these two expressions yields:
\[ f(x+y) + f(x-y) = \frac{a^{x+y} + a^{-(x+y)} + a^{x-y} + a^{-(x-y)}}{2}. \]
Using exponent properties, this simplifies to:
\[ f(x+y) + f(x-y) = \frac{a^x(a^y + a^{-y}) + a^{-x}(a^y + a^{-y})}{2}. \]
Factoring the expression gives:
\[ f(x+y) + f(x-y) = \frac{(a^x + a^{-x})(a^y + a^{-y})}{2}. \]
Recognizing that \( f(x) = \frac{a^x + a^{-x}}{2} \) and \( f(y) = \frac{a^y + a^{-y}}{2} \), we substitute these into the equation:
\[ f(x+y) + f(x-y) = 2f(x)f(y). \]
Final Answer:
\[ \boxed{2f(x)f(y)} \]