$6 \cdot 156$ Find all continuous functions $f:(1,+\infty) \rightarrow R$ satisfying
$$
f(x y) \equiv x f(y)+y f(x), x, y>1
$$
［Solution］First, prove that for any $k>0$, we have $f\left(x^{k}\right) \equiv k x^{k-1} f(x)$, where $x>1$.

The proof is divided into three steps:
(1) Let $k \in \mathbb{N}$. If $k=1$, then $f\left(x^{1}\right)=1 \cdot x^{0} \cdot f(x)$. Assume the identity holds for some $k \in \mathbb{N}$. Then,
$$
\begin{array}{r}
f\left(x^{k+1}\right) \equiv f\left(x^{k} \cdot x\right) \equiv x^{k} f(x) + x f\left(x^{k}\right) \\
\equiv x^{k} f(x) + x k x^{k-1} f(x) \equiv (k+1) x^{k} f(x).
\end{array}
$$

Thus, the identity holds for $k+1$. Therefore, the identity holds for any $k \in \mathbb{N}$.
(2) Let $k \in \mathbb{Q}, k>0$, i.e., $k=\frac{p}{q}$, where $p, q \in \mathbb{N}$. By (1),
$$
\begin{array}{l}
f\left(x^{p}\right) \equiv p x^{p-1} f(x) \\
f\left(\left(x_{q}^{p}\right)^{q}\right) \equiv q\left(x_{q}^{p}\right)^{q-1} f\left(x_{q}^{p}\right);
\end{array}
$$

The left sides of these two identities are equal, so we have
$$
f\left(x_{q}^{p}\right) \equiv \frac{p}{q} x_{q}^{p-1} f(x).
$$
(3) Let $k \in \mathbb{R}, k>0$, then take a sequence of positive rational numbers $k_{1}, k_{2}, \cdots$ such that
$$
\lim _{n \rightarrow \infty} k_{n}=k.
$$

Since the function $f(x)$ is continuous, for any $x>1$, we have
$$
f\left(x^{k}\right)=\lim _{n \rightarrow \infty} f\left(x^{k_{n}}\right)=\lim _{n \rightarrow \infty} k_{n} x^{k_{n}-1} f(x)=k x^{k-1} f(x).
$$

Next, using the proven identity, it is easy to find the expression for the function $f(x)$. Indeed, let $t = \ln x$, i.e., $x=e^{t}$, then we get
$$
f(x)=f\left(e^{t}\right)=t e^{t-1} f(e)=\ln x \cdot \frac{x}{e} f(e).
$$

On the other hand, for any $c \in \mathbb{R}$, the function $f(x)=c x \ln x$ satisfies the given conditions.