, 2022

Setting $y = \frac{t}{f(x)^3}$ we get
$$f(x+t) = x^3 f\left(\frac{t}{f(x)^3}\right) + f(x) \quad (1)$$
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 $c < 1$, taking $x = 1$ and $y = \frac{1}{1-c^3}$ we have $y - yc^3 = 1$, so $yf(1)^3 + 1 = y$ and $f(yf(1)^3 + 1) = f(y) = 1^3 f(y)$. Thus $f(1) = 0$, a contradiction. Assume now for contradiction that $c > 1$. We claim that
$$f(1 + c^3 + \cdots + c^{3n}) = (n+1)c$$
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 $x = 1 + c^3 + \cdots + c^{3n-3}$, $t = c^{3n}$ in (1) we get
$$(n+1)c = f(1 + c^3 + \cdots + c^{3n}) = (1 + c^3 + \cdots + c^{3n-3})f\left(\frac{c^{3n}}{(n+1)^3}\right) + nc$$
giving
$$f\left(\frac{c^{3n}}{(n+1)^3}\right) = \frac{c}{(1+c^3+\cdots+c^{3n})^3} < c = f(1) \implies \frac{c^{3n}}{(n+1)^3} < 1.$$
But this leads to a contradiction if $n$ is large enough. □

Now for $x = 1$ we get $f(y+1) = f(y)+1$ and since $f(1) = 1$ inductively we get $f(n) = n$ for every $n \in \mathbb{N}$. For $m, n \in \mathbb{N}$, setting $x = n, y = q = m/n$ we get
$$mn^2 + n = f(qn^3 + n) = f(yf(x)^3 + x) = x^3f(y) + f(x) = n^3f(q) + n \implies f(q) = q.$$
Since $f$ is strictly increasing with $f(q) = q$ for every $q \in \mathbb{Q}^{>0}$ we deduce that $f(x) = x$ for every $x > 0$. It is easily checked that this satisfies the functional equation.