Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Let $a \in \mathbb{R}_{>0}$ be given. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}$ such that $f(a)=1$ and

$$
\forall x, y \in \mathbb{R}_{>0}: f(x) f(y)+f\left(\frac{a}{x}\right) f\left(\frac{a}{y}\right)=2 f(x y)
$$
Substituting $x=a$ and $y=1$ into (1) gives $f(a) f(1)+f\left(\frac{a}{a}\right) f\left(\frac{a}{1}\right)=2 f(a \cdot 1)$, which, due to $f(a)=1$, leads to $f(1)+f(1)=2$, thus

$$
f(1)=1
$$

Substituting $y=1$ into (1) gives $f(x) f(1)+f\left(\frac{a}{x}\right) f\left(\frac{a}{1}\right)=2 f(x \cdot 1)$, which, due to $f(a)=f(1)=1$, leads to $f(x)+f\left(\frac{a}{x}\right)=2 f(x)$, thus

$$
f(x)=f\left(\frac{a}{x}\right) .
$$

But then (1) transforms into $f(x) f(y)+f(x) f(y)=2 f(x y)$, thus

$$
f(x) f(y)=f(x y)
$$

From (3) and (4) it follows that $f(x) f(x)=f(x) f\left(\frac{a}{x}\right)=f\left(x \cdot \frac{a}{x}\right)=f(a)=1$, so for every $x \in \mathbb{R}_{>0}$, $f(x)=1$ or $f(x)=-1$.
Since we can write positive $x$ as $\sqrt{x} \sqrt{x}$, we find using (4) that $f(x)=$ $f(\sqrt{x} \sqrt{x})=f(\sqrt{x}) f(\sqrt{x})=( \pm 1)^{2}=1$, thus $f(x)=1$ for every $x \in \mathbb{R}_{>0}$. We conclude that the only possibility for $f$ is apparently the constant function $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}: x \mapsto 1$.
Finally, we check whether the constant function $\forall x: f(x)=1$ indeed satisfies the conditions. The condition $f(a)=1$ is satisfied, while (1) becomes $1 \cdot 1+1 \cdot 1=2 \cdot 1$, which holds.
Thus, there is exactly one function $f$ that satisfies the given conditions, namely $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}: x \mapsto 1$.