6・167f is a function defined on $(1,+\infty)$ and takes values in $(1,+\infty)$, satisfying the condition: for any $x, y>1$ and $u, v>0$, we have
$$
f\left(x^{u} y^{v}\right) \leqslant f(x)^{\frac{1}{4 u}} f(y)^{\frac{1}{4 v}} .
$$

Determine all such functions $f$.
【Solution】Suppose $f$ satisfies the conditions in the problem, then we have
$$
f\left(x^{u} y^{v}\right) \leqslant f(x)^{\frac{1}{4 u}} f(y)^{\frac{1}{4 v}} .
$$

Taking $u=\frac{1}{2}$, we get
$$
f\left(x^{\frac{1}{2}} y^{v}\right) \leqslant f(x)^{\frac{1}{2}} f(y)^{\frac{1}{4 v}},
$$

Taking $v$ such that $\quad y^{v}=x^{\frac{1}{2}} \quad(x>1, y>1)$,
i.e.,
$$
v=\frac{\ln x}{2 \ln y} \text {. }
$$

Thus,
$$
\begin{array}{l}
f(x) \leqslant f(x)^{\frac{1}{2}} \cdot f(y)^{\frac{\ln y}{2 \ln x}}, \\
f(x)^{\frac{1}{2}} \leqslant f(y)^{\frac{\ln y}{2 \ln x}}, \\
f(x)^{\ln x} \leqslant f(y)^{\ln y} .
\end{array}
$$

By symmetry, we get $f(y)^{\ln y} \leqslant f(x)^{\ln x}$.
From (3) and (4), we have $f(x)^{\ln x}=f(y)^{\ln y}$.
This implies that $f(x)^{\ln x}$ is a constant $C(C>1)$, i.e.,
$$
\begin{array}{l}
f(x)^{\ln x}=C \text {, } \\
f(x)=C^{\frac{1}{\ln x}} \quad(C>1) \text {. } \\
\end{array}
$$

Conversely, if $f(x)=C^{\frac{1}{\ln x}}(C>1)$, then for $x, y>1, u, v>0$, we have
$$
\begin{aligned}
f\left(x^{u} y^{v}\right) & =C^{\frac{1}{\ln \left(x^{\prime \prime} y^{\prime \prime}\right)}} \\
& =C^{\frac{1}{u \ln x+v \ln y}} .
\end{aligned}
$$

Since $(u \ln x+v \ln y)^{2} \geqslant 4 u v \ln x \ln y$,
it follows that $\frac{1}{u \ln x+v \ln y} \leqslant \frac{u \ln x+v \ln y}{4 u v \ln x \ln y}$
$$
=\frac{1}{4 v \ln y}+\frac{1}{4 u \ln x} \text {. }
$$

Thus,
$$
\begin{aligned}
& f\left(x^{u} y^{v}\right) \\
= & C^{\frac{1}{4 \ln x+v \ln y}} \\
\leqslant & C^{\frac{1}{4 u \ln x}+\frac{1}{4 v \ln y}} \\
= & C^{\frac{1}{4 u \ln x}} \cdot C^{\frac{1}{4 v \ln y}} \\
= & f(x)^{\frac{1}{4 u}} \cdot f(y)^{\frac{1}{42}} .
\end{aligned}
$$

Therefore, the function that satisfies the conditions in the problem is
$$
f(x)=C^{\frac{1}{\ln x}} \quad(C>0) .
$$