I found this functional equation: $$xf(x)+x^2f(x-1)=f(x^2)\tag1$$ for all $x\in\mathbb{R}$. I tried to solve it (that is: find all functions $f:\mathbb{R}\to\mathbb{R}$ such that (1) is true); but I only found that $f(0)=0$ and classical substitution $x-1\mapsto x$ doesn't give me any information.
What do you think that can be a good approach?
Hint: (Assuming f(x) to be a polynomial)
Case 1: If f(x) is a constant function,you will get $f(x)=0$.
Case 2:If f(x) is a non constant function,it has to be a quadratic function(which you can realize by assuming f(x) to be of degree n and comparing the degrees of LHS and RHS).
Also,you can assume your quadratic to be $ax^2+bx$.(as the c in standard quadratic function $ax^2+bx+c$ will turn out to be $0$ if you plug $x=0$ in the equation.)