Problem. Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that

$$
f\left(y(f(x))^{3}+x\right)=x^{3} f(y)+f(x)
$$

for all $x, y>0$.
. Setting $y=\frac{t}{f(x)^{3}}$ we get

$$
f(x+t)=x^{3} f\left(\frac{t}{f(x)^{3}}\right)+f(x)
$$

for every $x, t>0$.
From (1) it is immediate that $f$ is increasing.
Claim. $f(1)=1$
Proof of Claim. Let $c=f(1)$. If $c<1$, then taking $x=1, t=1-c$ in (1) we get

$$
f(2-c)=f(1)=c
$$

which is a contradiction since $f$ is increasing. If $c>1$, we claim that

$$
f\left(1+c^{3}+\cdots+c^{3 n}\right)=(n+1) c
$$

for every $n \in \mathbb{N}$. We proceed by induction, the case $n=0$ being trivial. The inductive step follows easily by taking $x=1, t=c^{3}+c^{6}+\cdots+c^{3(k+1)}$ in (1).

Now taking $x=1+c^{3}+\cdots+c^{3 n-3}, t=c^{3 n}$ in (1) we get

$$
(n+1) c=f\left(1+c^{3}+\cdots+c^{3 n}\right)=\left(1+c^{3}+\cdots+c^{3 n-3}\right)^{3} f\left(\frac{c^{3 n}}{(c n)^{3}}\right)+n c
$$

giving

$$
f\left(\frac{c^{3 n-3}}{n^{3}}\right)=\frac{c}{\left(1+c^{3}+\cdots+c^{3 n}\right)^{3}}
$$

Since $c>1$, the right-hand side tends to 0 as $n \to \infty$, which contradicts the fact that $f(x)>0$ for every $x>0$. Therefore, $c=1$.

Now, we have $f(1)=1$. Taking $x=1$ in (1) we get

$$
f(1+t)=f\left(\frac{t}{f(1)^{3}}\right)+f(1) = f(t) + 1
$$

for every $t>0$. This implies that $f(x)=x$ for every $x>0$. It is easily checked that this satisfies the functional equation.