Konigsberg wrote: Deos there exist a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x$, $y \in \mathbb{R}$, $f(x^2y+f(x+y^2))=x^3+y^3+f(xy)$ Let $P(x,y)$ be the assertion $f(x^2y+f(x+y^2))=x^3+y^3+f(xy)$ The polynomial $(x^2+1)^2x+(x^2+1)x+f(-1)=0$ has degree $5$ and so always has at least one real root. Let then $u$ such a root. $P(-1-u^2,u)$ $\implies$ $(-1-u^2)^3+u^3=0$ and so $u^2+1=u$ which is impossible. So no solution for this functional equation.

Konigsberg wrote: Deos there exist a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x$, $y \in \mathbb{R}$, $f(x^2y+f(x+y^2))=x^3+y^3+f(xy)$ Another simpler way : Let $P(x,y)$ be the assertion $f(x^2y+f(x+y^2))=x^3+y^3+f(xy)$ $P(4,0)$ $\implies$ $f(f(4))=64+f(0)$ $P(0,2)$ $\implies$ $f(f(4))=8+f(0)$ So contradiction and no such function.

Let $P(x,y)$ be the given assertion, we have $P(0,x)\implies f(f(x))=x^3+f(0), P(x,0)\implies f(f(x^2))=x^3+f(0)$ and also we can see that $f$ is injective. Thus, $f(f(x))=f(f(x^2))\implies x=x^2 \ \forall x\in \mathbb{R}$ a contradiction and there is no such function.

Let $P(x,y):=f(x^2y+f(x+y^2))=x^3+y^3+f(xy)$ $P(x,0)$ yields $f(f(x))=x^3+f(0)$ thus $f$ is bijective since $x^3$ touches all real values $P(0,0)$ yields $f(f(0))=f(0)\overset{\text{bijectivity}}{\Longrightarrow}f(0)=0$ thus $P(x,0)$ transforms into $f(f(x))=x^3$ $P(0,x)$ yields $f(f(x^2))=x^3$ Using $P(x,0)\text{ and }P(0,x)$ yields $f(f(x))=f(f(x^2))\overset{\text{bijectivity}}{\Longrightarrow}f(x^2)=f(x)$ which forces $f$ to be constant which clearly contradicts bijectivity and $P(x,0)\text{ and }P(0,x)$ Therefore there exists no such function $\blacksquare$.