Here is my solution. Let $P(x,y)$ be the assertion. $P(0,0)$ gives $f(0)=0$. $P(x,0)$ implies $f(x)^3=x\,f(x^2)$ and $P(0,x)$ implies $f(x)^3=f(f(x))\,x^2$ for every $x\in\mathbb{R}$. So, $f(x)^3=x\,f(x^2)=x^2f(f(x))$. Since it is already true for $x=0$, we can say that $f(x^2)=x\,f(f(x))$ for every $x\in\mathbb{R}$ and replace in the original equation to get $$f(x+y)^3=(x^2+2xy)f(f(x))+(x^2+3xy+y^2)f(f(y))$$Reversing the $x,y$ order, we get $$f(x+y)^3=(x^2+3xy+y^2)f(f(x))+(2xy+y^2)f(f(y))$$which means $$f(f(x))(x+y)y=f(f(y))(x+y)x$$setting $y=1$ and letting $f(f(1))=k$, we have $f(f(x))=kx$ for every $x\in\mathbb{R}-\{-1\}$. So, for $x,y\in\mathbb{R}-\{-1\}$, $f(x+y)^3=k(x^3+3x^2y+3xy^2+y^3)$, and $f(x+y)=(x+y)t$ where $t=\sqrt[3]{k}$. This means that $f(x)\equiv tx$ for a constant $t$. Putting this into $f(x^2)=x\,f(f(x))$, we get $t=t^2$, so $t=0$ or $t=1$. Therefore, the solutions are $f(x)=0$ and $f(x)=x$.

\[f(x+y)^3=(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)\]$P(0,0)\implies f(0)^3=0\iff f(0)=0$ $P(0,y)\implies f(y)^3=f(f(y)).y^2$ $P(x,0)\implies f(x)^3=xf(x^2)$ Hence $f(f(x)).x^2=f(x)^3=xf(x^2)\iff x.f(f(x))=f(x^2)$ $P(x,y)=P(y,x)\implies (x+2y)f(x^2)+(x^2+3xy+y^2).f(f(y))=(y+2x)f(y^2)+(x^2+3xy+y^2).f(f(x))$ $(x+y)(f(x^2)-f(y^2))+yf(x^2)-xf(y^2)+(x^2+3xy+y^2)(f(f(y))-f(f(x)))=0$ $P(-2y,y)\implies f(-y)^3=f(f(y)).(4y^2-6y^2+y^2)=f(f(y)).-y^2$ Also $-f(f(y)).y^2=f(-y)^3=-y.f(y^2)\implies f(y^2)=y.f(f(y))$ Thus $\boxed{f(f(x))=\frac{f(x^2)}{x}}$ \[f(x+y)^3=(x+2y).f(x^2)+\frac{f(y^2)}{y}(x^2+3xy+y^2)=(x+2y).f(x^2)+(2x+y).f(y^2)+(x+\frac{x^2}{y}).f(y^2)\]\[f(x+y)^3-(x+2y).f(x^2)-(y+2x).f(y^2)=(x+\frac{x^2}{y}).f(y^2)\]While swapping $x\leftrightarrow y$, $LHS$ doesn't change so $(x+\frac{x^2}{y}).f(y^2)=(y+\frac{y^2}{x}).f(x^2)$ $\iff (x^2y+x^3).f(y^2)=(xy^2+y^3).f(x^2)$ $\iff (x+y).x^2f(y^2)=(x+y).y^2f(x^2)$ $\iff (x+y)(\frac{f(x^2)}{x^2}-\frac{f(y^2)}{y^2})=0$ $x+y\neq 0\implies \frac{f(x^2)}{x^2}=\frac{f(y^2)}{y^2}$ Taking $x,y>0$ gives that $\boxed{f(x^2)=cx^2}$ Also by taking $x+y<0,x>0$ we get that $c=\frac{f(x^2)}{x^2}=\frac{f(y^2)}{y^2}$ So $f(x^2)=cx^2$ for all $x\in R$. $c^3x^6=f(x^2)^3=x^2.f(x^4)=cx^6\implies c=c^3\iff \boxed{c=-1,0,1}$ $f(f(x^2))=\frac{f(x^4)}{x^2}=cx^2$ If $c=-1,$ then $x^2=-x^2$ gives contradiction. If $c=0\iff f\equiv 0$ for all $x>0,$ then $f(x+y)^3=0\implies \boxed{f\equiv 0, \forall x\in R}$ If $c=1 \iff f(x)=x$ for all $x>0,$ then $f(x)^3=x.f(x^2)=x^3\implies \boxed{f(x)=x, \forall x\in R}$

Let $P(x,y)$ denote the assertion $f(x+y)^3=(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)$ First, by $P(0,0)$ we have $f(0)=0$ By $P(0,x)$ and by $P(x,0)$ we can also deduce that $f(x)^3=xf(x^2)=f(f(x))\cdot x^2$. Plugging in $x\rightarrow -x$ we have $f$ is odd. Also let's keep in mind that $f(f(x))=\frac{f(x^2)}{x}$ for any $x\neq 0$ Swapping x and y we get that $xf(y^2)+x^2\cdot f(f(y))=y^2\cdot f(f(x))+yf(x^2)$ and using $f(f(x))=\frac{f(x^2)}{x}$ with $x,y\neq 0$ we have $xf(y^2)+\frac{f(y^2)\cdot x^2}{y}=yf(x^2)+\frac{f(x^2)\cdot y^2}{x} $. Plugging $x=1$ in the last one for $y\neq0,-1$ we get that $f(y^2)=y^2f(1)$ so $f(x)=xf(1)$ for any $x\geq 0$. Using f odd we get $f(x)=xf(1)$ for any $x\in \mathbb R$ and we are done

My solution is same with #4

Here is my solution Let $P(x,y)$ denote the assertion. $P(0,0)$ gives $f(0)=0$. $P(x,0)$ gives $f(x)^3=xf(x^2)$ and $P(-x,0)$ gives $f(-x)^3=(-x)f(x^2)$, so $f$ is odd function. $P(x,-x)$ gives $f(x^2)=xf(f(x))$. Now $P(x,y)$ and $P(y,x)$ gives that $$(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)=f(x+y)^3=(y+2x)f(y^2)+f(f(x))(x^2+3xy+y^2)$$and if we put $xf(f(x))$ instead of $f(x^2)$ we get that $$(x+2y)xf(f(x)) +f(f(y))(x^2+3xy+y^2)= (y+2x)yf(f(y))+f(f(x))(x^2+3xy+y^2)$$, so we get $f(f(x))y(x+y)=f(f(y))x(x+y)$, so $\frac {f(f(x))}{x}$ is constant for all $x\in\mathbb{R}$, so $f(f(x))=cx$ and $f(x^2)=cx^2$ from $f(x^2)=xf(f(x))$. And we put this in the condition then we get that $f(x)=ax$. Again we put this in the condition then we get that $a=0$ or $a=1$. Therefore, the solutions are $f(x)=0$ and $f(x)=x$.

Let the given assertion be $P(x,y)$. The only solutions are $\boxed{f(x)=0, f(x)=x}$ which can both be easily verified. Claim: $f(0)=0$ This is immediate from $P(0,0)$ Claim: $f(x)$ is odd Below are the assertions $P(-2y,y)$ and $P(0,y)$, $$f(-y)^3=-y^2f(f(y))$$$$f(y^3)=y^2f(f(y)$$The result follows. Claim: $f(x^2)=xf(f(x))$ This follows from $P(x,-x)$ combined with the fact that $f(x)$ is odd. Claim: $f(x)=cx$ for all $x\in \mathbb{R}^{+}$ Consider $P(x,y)$ and $P(y,x)$ we get, $$(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)=(y+2x)f(y^2)+f(f(x))(x^2+3xy+y^2)$$Multiplying by $xy$ and then using the reduction shown in the previous claim we get, $$(xy^2+y^3)f(x^2)=(yx^2+x^3)f(y^2)$$If for any $x>0$ we have $f(x)=0$ then $f(x)=0$ for all $x\in \mathbb{R}^{+}$. Otherwise we get $\frac{f(x^2)}{x^2}=\frac{f(y^2)}{y^2}$ for all $x,y\in \mathbb{R}^{+}$ which implies the result. Claim: Either $f(x)=x$ or $f(x)=0$ Given that $f$ is odd and that $f(0)=0$ the previous claim implies that $f(x)=cx$ for all $x$. As $f(x^2)=xf(f(x))$ we must have $c=c^2$ so either $c=0$ or $c=1$.

$P(0,0) \implies f(0) = 0, P(x,0) \implies f(x)^3 = xf(x^2)$* and $x \rightarrow -x \implies -f(x) = f(-x)$. $P(-2y,y) \implies f(f(y)) = f(y)^3/y^2 \overset{*} = f(y^2)/y$. Thus, $$f(x+y)^3 = xf(x^2) + 2yf(x^2) + \frac{x^2}{y} f(y^2) + 3xf(y^2) + yf(y^2) = xf(x^2) + 2xf(y^2) +\frac{y^2}{x} f(x^2) + 3yf(x^2) + yf(y^2) \implies $$$$xf(y^2) + \frac{x^2}{y} f(y^2) = yf(x^2) + \frac{y^2}{x} f(x^2) \overset{y \rightarrow 1} \implies f(x^2) = f(1) x^2$$Because $f$ is odd, $f(x) = cx$ and plugging back gives $f \equiv x$ or $f \equiv 0$.

Solution sketch : $P(0,0)$ gives $f(0)=0$ $P(0,x)$ and $P(x,0)$ gives $f(x)^3=x^2f(f(x))=xf(x^2)$. $P(1,y)-P(y,1)$ : $f(f(y))+\frac{f(f(y))}{y}=c+cy$ where $f(f(1))=c$ $\Rightarrow f(f(y))=cy$. Thus $f(y)=c^{\frac{1}{3}}y$ We find then that $c=0,1$ which finishes the problem.