6-14 The function $f$ defined on ordered pairs of positive integers satisfies the following three properties:
(1) $f(x, x)=x$;
(2) $f(x, y)=f(y, x)$;
(3) $(x+y) f(x, y)=y f(x, x+y)$.

Try to compute: $f(14,52)$.
[Solution] From (3), we get
$$f(x, x+y)=\frac{x+y}{y} \cdot f(x, y)$$

Thus, when $z>x$, we have
$$f(x, z)=\frac{z}{z-x} f(x, z-x) .$$

Therefore,
$$\begin{aligned}
f(14,52) & =\frac{52}{38} f(14,38) \\
& =\frac{52}{38} \cdot \frac{38}{24} f(14,24) \\
& =\frac{52}{24} \cdot \frac{24}{10} f(14,10) \\
& =\frac{52}{10} f(10,14) \\
& =\frac{52}{10} \cdot \frac{14}{4} f(10,4) \\
& =\frac{91}{5} f(4,10) \\
& =\frac{91}{5} \cdot \frac{10}{6} \cdot f(4,6) \\
& =\frac{91}{3} \cdot \frac{6}{2} f(4,2) \\
& =91 \cdot f(2,4) \\
& =91 \cdot \frac{4}{2} f(2,2) \\
& =91 \cdot 2 \cdot 2 \\
& =364 .
\end{aligned}$$