Solution:
Put $x=0$ and we get $f(y f(0))=0$. If $f(0) \neq 0$, then $y f(0)$ takes all real values when $y$ varies over real line. We get $f(x) \equiv 0$. Suppose $f(0)=0$. Taking $y=-x$, we get $f\left(x^{2}-x f(x)\right)=0$ for all real $x$.

Suppose there exists $x_{0} \neq 0$ in $\mathbb{R}$ such that $f\left(x_{0}\right)=0$. Putting $x=x_{0}$ in the given relation we get
$$f\left(x_{0}^{2}\right)=x_{0} f\left(x_{0}+y\right)$$
for all $y \in \mathbb{R}$. Now the left side is a constant and hence it follows that $f$ is a constant function. But the only constant function which satisfies the equation is identically zero function, which is already obtained. Hence we may consider the case where $f(x) \neq 0$ for all $x \neq 0$.

Since $f\left(x^{2}-x f(x)\right)=0$, we conclude that $x^{2}-x f(x)=0$ for all $x \neq 0$. This implies that $f(x)=x$ for all $x \neq 0$. Since $f(0)=0$, we conclude that $f(x)=x$ for all $x \in \mathbb{R}$.
Thus we have two functions: $f(x) \equiv 0$ and $f(x)=x$ for all $x \in \mathbb{R}$.