20.15. (Austria, 75). Find all continuous functions $\mathrm{f}:(1 ;+\infty) \rightarrow \mathbf{R}$, satisfying the identity

$$
f(x y) \equiv x f(y)+y f(x), \quad x, y>1
$$
20.15. We will prove that for any value of $k>0$ the identity

$$
f\left(x^{k}\right) \equiv k x^{k-1} f(x) \quad(x>1)
$$

is valid. The proof will be conducted in three stages.

1) Let $k \in \mathbb{N}$. If $k=1$, then we have

$$
f\left(x^{1}\right)=1 \cdot x^{0} \cdot f(x)
$$

and if the identity is valid for some value of $k \in \mathbb{N}$, then it is also valid for the value $k+1$, since

$f\left(x^{k+1}\right)=f\left(x^{k} x\right) \equiv x^{k} f(x)+x f\left(x^{k}\right)=$

$$
x^{k} f(x)+x k x^{k-1} f(x)=(k+1) x^{k} f(x)
$$

By the principle of mathematical induction, the identity is valid for any $k \in \mathbb{N}$.

2) Let $k \in \mathbb{Q}, k>0$, i.e., $k=p / q$, where $p, q \in \mathbb{N}$. From what was proven in point 1), we have two identities

$$
\begin{gathered}
f\left(x^{p}\right)=p x^{p-1} f(x), \\
f\left(\left(x^{p / q}\right)^{q}\right)=q\left(x^{p / q}\right)^{q-1} f\left(x^{p / q}\right) .
\end{gathered}
$$

Equating the right-hand sides of these identities, we get

$$
f\left(x^{p / q}\right) \equiv(p / q) x^{p / q-1} f(x)
$$

i.e., the identity is valid for any positive rational $k$.

3) Let $k \in \mathbb{R}, k>0$. Then choose a sequence of positive rational numbers $k_{1}, k_{2}, \ldots$, such that

$$
\lim _{n \rightarrow \infty} k_{n}=k
$$

Since the function $f(x)$ is continuous, for any value of $x>1$ we have

$$
f\left(x^{k}\right)=\lim _{n \rightarrow \infty} f\left(x^{k_{n}}\right)=\lim _{n \rightarrow \infty} k_{n} x^{k_{n}-1} f(x)=k x^{k-1} f(x) .
$$

From the proven identity, it is easy to find the explicit form of the function $f(x)$. Indeed, denoting $t=\ln x$, i.e., $x=e^{t}$, 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 value of $c \in \mathbb{R}$, the function $f(x)=c x \ln x$ satisfies the condition of the problem.