It is easy to see that $f(0) = 0$ and $f(x^2) = x f(x)$ for all $x \in \mathbb{R}$. If $f(\alpha) = 0$ for some $\alpha$, then
$$f(x^2 + x f(\alpha)) = x f(x + \alpha),$$
for all $x \in \mathbb{R}$. Therefore
$$x f(x + \alpha) = f(x^2) = x f(x),$$
for all $x \in \mathbb{R}$. For $x \neq 0$, we get $f(x) = f(x + \alpha)$. Note that this is also valid for $x = 0$. Suppose $f(1) = 0$. Then
$$f(1 + f(x)) = f(x + 1),$$
so that $f(f(x)) = f(x)$ for all $x$. If there exists a $\lambda \in \mathbb{R}$ such that $f(\lambda) \neq 0$, then for any $t \in \mathbb{R}$, taking $s = t / f(\lambda)$, we have
$$f(s^2 + s f(\lambda - s)) = s f(s + \lambda - s) = s f(\lambda) = t$$
which shows that $f$ is onto. Hence there exists $x_0$ such that $f(x_0) = 1$. This gives
$$1 = f(x_0) = f(f(x_0)) = f(1) = 0,$$
which is absurd. Thus $f(1) = 0$ forces $f(x) = 0$ for all $x \in \mathbb{R}$.

Suppose $f(1) \neq 0$. Let $\alpha \in \mathbb{R}$ be such that $f(\alpha) = 0$. As we have seen earlier, $f(x + \alpha) = f(x) = 0$ for all $x$. Therefore
$$\begin{aligned} f(\alpha^2 + 1) &= f((1 + \alpha)^2 + (1 + \alpha) f(\alpha)) = (\alpha + 1) f(1 + 2\alpha) = (\alpha + 1) f(1), \\ f(\alpha^2 + 1) &= f((1 - \alpha)^2 + (1 - \alpha) f(\alpha)) = (1 - \alpha) f(1). \end{aligned}$$
These show that $1 + \alpha = 1 - \alpha$. Therefore $\alpha = 0$. Thus $f(\alpha) = 0$ implies that $\alpha = 0$.

Taking $x = -y$ in the equation, we get $f(y^2 - y f(y)) = y f(y - y) = 0$. Hence $y^2 - y f(y) = 0$ for all $y \in \mathbb{R}$. This gives $f(y) = y$ for all $y \neq 0$. Since $f(0) = 0$, we conclude that $f(y) = y$ for all $y \in \mathbb{R}$.