This problem was proposed by Greece (Athanasios Kontogeorgis, aka socrates) at the BMO SL 2016.

IMO2019 wrote: Let, $P(x,y)$ be assertion. Also, let there is $k\in \mathbb{R}-\{0\}$ such that $f(k)=0,f(0)\not=0$. Then for $P(k,0)$. $$f(f(k)+0)=f(k)+3.0+0.f(0)\implies f(0)=0$$Contradiction. So we have, $f(0)=0$. For $P(0,x)$ $$f(x)=x.f(x)\implies f(x)=0$$which doesn't suffice the given condition so there are no solutions. This in an equation from $\mathbb{R}^+$ to itself (so plugging $0$ is not allowed, unfortunately). IstekOlympiadTeam wrote: Prove that there does not exist a function $f : \mathbb R^+\to\mathbb R^+$ such that \[f(f(x)+y)=f(x)+3x+yf(y)\]for all positive reals $x,y$. This is longer than the other solution, but I still think it works We are given $f(f(x)+y)=f(x)+3x+yf(y)$ for the sake of reference. Let $P(x,y)$ be this assertion. $P(x,y)-P(x,z)$ gives $f(f(x)+y)-f(f(x)+z)=yf(y)-zf(z)$. Let $S=\text{Range}(f)$ (clearly $|S|\geq 1$). Suppose that $|S|\geq 2$, and pick distinct $v,v_0\in S$ (WLOG $v>v_0$) with $f(u)=v$ and $f(u_0)=v_0$. $P(u,y)-P(u,z)$: $f(v+y)-f(v+z)=yf(y)-zf(z)$ $P(u_0,y)-P(u_0,z)$: $f(v_0+y)-f(v_0+z)=yf(y)-zf(z)$ Combining these two equations yields $f(y+v)-f(z+v)=f(y+v_0)-f(z+v_0)$. Hence there exists some real $c$ (possibly negative) such that $f(y+v)=f(y+v_0)+c$. Shifting $y\to y-v_0$ and letting $t=v-v_0$ gives $f(y+t)=f(y)+c$ for all $y>v_0$. Suppose $y\geq v_0$. Then $P(x,y+t)$ yields $f(f(x)+y+t)=f(x)+3x+(y+t)f(y+t)$, or \[f(f(x)+y)+c=f(x)+3x+(y+t)(f(y)+c).\]Subtracting $P(x,y)$ from this yields \[c=(y+t)(f(y)+c)-yf(y)=tf(y)+c(y+t)\ \Leftrightarrow\ f(y)=\frac{c(1-y-t)}{t}\ \forall y\geq v_0.\]However, notice that $c$ must be positive, because \[f(y+t)=f(y)+c\ \Rightarrow\ f(y+nt)=f(y)+nc\ \forall y\geq v_0\]and setting $n$ large enough gets a contradiction if $c<0$. But in this case, you have $f(y)=\frac{c(1-y-t)}{t}\leq 0$ for $y$ large, which is a contradiction anyway. It remains to plug in the constant function but this trivially does not work.

IstekOlympiadTeam wrote: Prove that there does not exist a function $f : \mathbb R^+\to\mathbb R^+$ such that \[f(f(x)+y)=f(x)+3x+yf(y)\]for all positive reals $x,y$. Let $P(x,y)$ be the assertion $f(f(x)+y)=g(x)+yf(y)$ where $g(x)=f(x)+3x$ $P(x,y+f(z))$ $\implies$ $f(f(x)+f(z)+y)=g(x)+(y+f(z))f(f(z)+y)$ which is : $f(f(x)+f(z)+y)=g(x)+(y+f(z))(g(z)+yf(y))$ Switching there $x,z$ and subtracting, we get : $yf(y)(f(z)-f(x))=y(g(x)-g(z))+(f(x)g(x)-f(z)g(z)+g(z)-g(x))$ And since no constant solution exists, just choosing $f(z)\ne f(x)$, this becomes $f(y)=a+\frac by$ for some $a,b$ And it is easy to check back that this is never a solution. Q.E.D.

Here's another solution different from the above: Suppose such a function exists. First,suppose there is a number $x$ in the domain of the function such that $x>f(x)$.Then plugging $y=x-f(x)$ gives that: $0=3x+(x-f(x)f(x-f(x))$,obviously impossible because we're dealing with strictly positive numbers. So,it turns out that $f(x)\geq x$ for every $x$ in the function's domain. Now plugging in $y=f(y)$ and exploiting the symmetry created we get: $f(f(x))f(x)-f(x)=3x+c$ for all $x$ in the domain and for some real constant $c$. We will prove that this cannot hold for very large $x$. Indeed,choose some huge number $x$ such that both the RHS and LHS are strictly positive.Then we would have: $x\leq f(x)=\frac{3x+c}{f(f(x))-1}\leq \frac{3x+c}{x-1}$,which implies that $x^2-4x\leq c$. However,since $x$ can be made arbitrarily large,this is obviously a contradiction.

Similar to above.