China Girls' Mathematical Olympiad

Setting $x = y = 1$ in $(1)$ gives
$$f^2(1) + f^2(a) = 2f(1),$$
$$(f(1) - 1)^2 = 0,$$
so $f(1) = 1$.

Setting $y = 1$ in $(1)$ yields
$$f(x)f(1) + f(\frac{a}{x})f(a) = 2f(x),$$
$$f(x) = f(\frac{a}{x}), \quad x > 0. \qquad (2)$$

Setting $y = \frac{a}{x}$ in $(1)$ yields
$$f(x)f(\frac{a}{x}) + f(\frac{a}{x})f(x) = 2f(a),$$
$$f(x)f(\frac{a}{x}) = 1. \qquad (3)$$

Combining $(2)$ and $(3)$ gives $f^2(x) = 1, x > 0$.

Setting $x = y = \sqrt{t}$ in $(1)$ gives
$$f^2(\sqrt{t}) + f^2(\frac{a}{\sqrt{t}}) = 2f(t),$$
$$f(t) > 0.$$
So $f(x) = 1, x > 0$, as desired.