13 Given non-zero real numbers $a$ and $b$, find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that
$$
\begin{aligned}
f(2 x) & =a f(x)+b x, \\
f(x) f(y) & =f(x y)+f\left(\frac{x}{y}\right)
\end{aligned}
$$

for all $x \in \mathbf{R}$ and all $y \in \mathbf{R}, y \neq 0$. Here $\mathbf{R}$ denotes the set of real numbers.
Assume $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies (1) and (2).
For $x \neq 0$, from (2) we know
$f(x) f(1)=f(x)+f(x) . \quad f(1) f(x)=f(x)+f\left(\frac{1}{x}\right)$.
Therefore, $f(x)=f\left(\frac{1}{x}\right)$ holds for any $x \neq 0$.
Also, from $f\left(\frac{2}{x}\right)=a f\left(\frac{1}{x}\right)+\frac{b}{x}$ we know
$$
f\left(\frac{x}{2}\right)=a f(x)+\frac{b}{x}=a\left(a f\left(\frac{x}{2}\right)+b \frac{x}{2}\right)+\frac{b}{x} \text {. }
$$

Thus, $\left(1-a^{2}\right) f\left(\frac{x}{2}\right)=\frac{1}{2} a b \cdot x+\frac{b}{x}$.
Clearly, $1-a^{2} \neq 0$, so $f\left(\frac{x}{2}\right)=\frac{a b}{2\left(1-a^{2}\right)} x+\frac{b}{1-a^{2}} \frac{1}{x}$. Therefore,
$$
f(x)=\frac{a b}{1-a^{2}} x+\frac{b}{2\left(1-a^{2}\right)} \frac{1}{x},
$$

Substituting into (1) gives
$$
\frac{a b}{1-a^{2}} 2 x+\frac{b}{4\left(1-a^{2}\right)} \frac{1}{x}=\frac{a^{2} b}{1-a^{2}} x+\frac{a b}{2\left(1-a^{2}\right)} \frac{1}{x}+b x .
$$

For any $x$ to hold, we know $a=\frac{1}{2}$.
Thus,
$$
f(x)=\frac{2 b}{3}\left(x+\frac{1}{x}\right) .
$$

From (2), we know $f(1)^{2}=2 f(1)$. From (3), we know $f(1)=\frac{4 b}{3}$. Therefore, $b=\frac{3}{2}$. Hence,
$$
f(x)=x+\frac{1}{x}(x \neq 0) \text {. }
$$

From (2), we have $f(0) f(y)=2 f(1))(y \neq 0)$, thus $f(0)=0$.
Therefore, $f(x)=\left\{\begin{array}{ll}x+\frac{1}{x}, & x \neq 0, \\ 0, & x=0 .\end{array}\right.$
When $a=\frac{1}{2}, b=\frac{3}{2}, f(x)=\left\{\begin{array}{ll}x+\frac{1}{x} & x \neq 0 \\ 0 & x=0\end{array}\right.$, we have
$$
f(2 x)=\left\{\begin{array}{ll}
2 x+\frac{1}{2 x}, & x \neq 0, \\
0, & x=0 ;
\end{array}\right.
$$
$a f(x)+b x=\frac{1}{2} f(x)+\frac{3}{2} x$
$$
=\left\{\begin{array}{ll}
\frac{1}{2}\left(x+\frac{1}{x}\right)+\frac{3}{2} x, x \neq 0 \\
0, & x=0
\end{array}=\left\{\begin{array}{ll}
2 x+\frac{1}{2 x}, & x \neq 0, \\
0 . & x=0 .
\end{array}\right.\right.
$$

Thus, $f(2 x)=a f(x)+b x$.
When $x=0, y \neq 0$, $f(x) f(y)=f(0) f(y)=0, f(x y)+$ $f\left(\frac{x}{y}\right)=2 f(0)=0$;

When $x \neq 0, y \neq 0$, $f(x) f(y)=\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)=x y+$ $\frac{y}{x}+\frac{x}{y}+\frac{1}{x y}=f(x y)+f\left(\frac{x}{y}\right)$.
Therefore, $f(x) f(y)=f(x y)+f\left(\frac{x}{y}\right)(y \neq 0)$.
In summary, we have the following conclusion:
When $a=\frac{1}{2} \cdot b=\frac{3}{2}$, the required $f$ is $f(x)=\left\{\begin{array}{ll}x+\frac{1}{x}, & x \neq 0, \\ 0 . & x=0 ;\end{array}\right.$; when $a \neq \frac{1}{2}$ or $b \neq \frac{3}{2}$, $f$ does not exist.