Determine all $f : \mathbb R \rightarrow \mathbb R$ such that $f(x^3 + y^3) = x^2f(x) + yf(y^2)$. 
Determine all $f : \mathbb R \rightarrow \mathbb R$ such that $$f(x^3 + y^3) = x^2\,f(x) + y\,f(y^2)$$
  for all $x,y\in\mathbb{R}$.

I put $x = y = 0$ then $x = 0$ and $y = 0$ respectively. This implies Cauchy’s Functional Equation , but I can’t continue after this.
 A: By setting $x$ and $y$ to be $0$, we obtain $f(0)=0$.  Now, with $y:=0$, we get
$$f(x^3)=x^2\,f(x)\tag{*}$$
for all $x\in\mathbb{R}$.  Likewise, with $x:=0$, we get
$$f(y^3)=y\,f(y^2)\tag{#}$$
for all $y\in\mathbb{R}$.  This shows that
$$x\,f(x^2)=f(x^3)=x^2\,f(x)$$
for every $x\in\mathbb{R}$.  Thus, along with $f(0)=0$, we conclude that
$$f(x^2)=x\,f(x)\tag{$\star$}$$
for all $x\in\mathbb{R}$. 
Now, we have by (*) and (#) that
$$\begin{align}f(x+y)&=f\Big((\sqrt[3]{x})^3+(\sqrt[3]{y})^3\Big)\\&=(\sqrt[3]{x})^2\,f\big(\sqrt[3]{x}\big)+\sqrt[3]{y}\,f\big((\sqrt[3]{y})^2\big)\\&=f\big((\sqrt[3]{x})^3\big)+f\big((\sqrt[3]{y})^3\big)=f(x)+f(y)\tag{\$}\end{align}$$
for every $x,y\in\mathbb{R}$.  From ($\star$) and (\$), we have
$$f\big((x+1)^2\big)=(x+1)\,f(x+1)=(x+1)\,\big(f(x)+f(1)\big)$$
for any $x\in\mathbb{R}$.  However, from ($\star$) and (\$), we also have
$$\begin{align}f\big((x+1)^2\big)&=f(x^2+2x+1)\\&=f(x^2)+2\,f(x)+f(1)=x\,f(x)+2\,f(x)+f(1)\end{align}$$
for any $x\in\mathbb{R}$.  That is,
$$(x+1)\,\big(f(x)+f(1)\big)=x\,f(x)+2\,f(x)+f(1)$$
for any $x\in\mathbb{R}$, whence
$$f(x)=f(1)\,x$$
for all $x\in\mathbb{R}$.
