BMO 2022 shortlist

Setting $y=\frac{t}{f(x{)}^3}$ we get $$f(x+t)=x^{3}f\left(\frac{t}{f(x{)}^3}\right)+f(x)(1)$$ 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 $y-y{c}^3=1$, 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 $$f\left(\frac{c^{3n}}{(n+1{)}^3}\right)=\frac{c}{(1+c^3+\cdots +c^{3n}{)}^3}<c=f(1)⟹\frac{c^{3n}}{(n+1{)}^3}<1.$$ 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 $$m{n}^2+n=f(q{n}^3+n)=f(yf(x{)}^3+x)=x^{3}f(y)+f(x)=n^{3}f(q)+n⟹f(q)=q.$$ 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.