Let $\rm P(x,y): f(x^2)f(y^2) + |x|f(-xy^2) = 3|y|f(x^2y) $. $\rm P(0,0): f(0)=0$, so we consider below that $\rm x,y\neq 0$. $\rm P(-x,y): f(x^2)f(y^2) + |x|f(xy^2) = 3|y|f(x^2y)$ so we have that $\rm |x|f(xy^2)=|x|f(-xy^2)\Leftrightarrow f(x)=f(-x)$. So we work in interval $\rm (0,+\infty)$. $\rm P(y,x): f(x^2)f(y^2)+yf(yx^2)=3xf(y^2x)$ so we have that $\rm xf(xy^2)-yf(yx^2)=3yf(x^2y)-3xf(y^2x)\Leftrightarrow xf(xy^2)=yf(x^2y)\,\,\,(*)$.From this for $\rm y=1$ we have $\rm xf(x)=f(x^2)$. Let $\rm f(1)=a$.$\rm P(1,1): a^2+=3a \Leftrightarrow a=0\,\,or \,\,\,a=2$. If $\rm a=0$ then $\rm P(x,1): xf(x)=3f(x^2)\Leftrightarrow 3f(x^2)=f(x^2)\Leftrightarrow f(x^2)=0\Leftrightarrow f(x)=0,x>0$. Since $\rm f(x)=f(-x)$ we have $\boxed {\rm f(x)=0,\forall x\in \mathbb{R}}$. If $\rm a=2$ we set $\rm x=\dfrac{1}{y^2}$ at $(*)$ : $\rm \dfrac{2}{y^2}=yf(\dfrac{1}{y^3})\overset{\dfrac{1}{y^3}\rightarrow x }{\Leftrightarrow} f(x)=2x$.Since $\rm f(x)=f(-x)$ we have $\boxed {\rm f(x)=|2x|,\forall x\in \mathbb{R}}(\rm verified)$

The date doesn't fit name of the forum. 2019 is old?

My posting ability in public forums is as everyone’s else limited. Therefore when I post in public forums, I post only the geometry problems. When my source is written not in English, only the geometry ones are posted as I spend time translating them by google translate. When my source is written in English, I post the whole problem set in aops, posting the non-geometry problems in this forum and adding all the problems in contest collections. In this forum’s index, anyone may find which contests’ years have been posted inside this forum.

Now I get it.

Fun problem ! Let $P(x,y) : f(x^2)f(y^2) + |x|f(-xy^2) = 3|y|f(x^2y)$. $P(-x,y) : f(x^2)f(y^2)+|-x|f(-xy^2)=3|y|f(x^2y)$ . ( ) $P(x,y) : f(x^2)f(y^2) + |x|f(-xy^2) = 3|y|f(x^2y)$ . ( ) ( ) - ( ) then we can see $|-x|f(xy^2)=|x|f(-xy^2) $ . If $y=1$ we can see $\left | \frac{-x}{x} \right | = \frac{f(-x)}{f(x)}$ or $f(x)=f(-x)$ . $P(y,x) : f(x^2)=|x|f(x)$ . ( ) $P(x,1) : |x|f(1)f(x)+|x|f(x)=3f(x^2) \to |x|f(1)f(x)+f(x^2)=3f(x^2) \to f(1)=2$ By ( ) and $P(\frac{1}{x^2} , x) : \frac{2}{y^2} = yf(\frac{1}{y^3})$ we get $f(x)=|2x|$ . $\blacksquare$

Here is my solution: https://calimath.org/pdf/EstoniaTST2019-3.pdf And I uploaded the solution with motivation to: https://youtu.be/bOmrlmG1Gig