The solutions are $f(x) = 0$ for all $x \in \mathbb{R}$ and $f(x) = x$ for all $x \in \mathbb{R}$.

It is easy to check that both are solutions. Label the equation as follows.
$$f(x + y f(x)) = f(x) + x f(y) \tag{1}$$
Putting $x = 1$ in (1), we have
$$f(1 + y f(1)) = f(1) + f(y). \tag{2}$$
If $f(1) \neq 1$, then there exists $y \in \mathbb{R}$ such that $1 + y f(1) = y$. For this $y$, we obtain $f(1) = 0$. Then (2) becomes $f(y) = 0$ for all $y$, which is the first solution.
Now, we assume $f(1) = 1$. If $f(t) = 0$ for some $t \in \mathbb{R}$, by putting $x = t$ and $y = 1$ in (1), we obtain $t = 0$. This means $f(x) \neq 0$ for any $x \neq 0$. Putting $x = 1$ in (1), we have
$$f(y + 1) = f(y) + 1. \tag{3}$$
By induction we easily obtain $f(n) = n$ for any $n \in \mathbb{Z}$. Putting $x = -1$ in (1), we find that
$$f(-y - 1) = -f(y) - 1.$$
Using (3), we get $-f(y) = f(-y - 1) + 1 = f(-y)$. Replacing $y$ by $-y$ in (1), we have
$$f(x - y f(x)) = f(x) - x f(y).$$
Adding this to (1), we have
$$f(x + y f(x)) + f(x - y f(x)) = 2 f(x).$$
For $x \neq 0$, we can replace $y$ by $\frac{y}{f(x)}$. This yields
$$f(x + y) + f(x - y) = 2 f(x). \tag{4}$$
Note that this also holds when $x = 0$. Considering $x = y$, we find that
$$f(2x) = 2 f(x).$$
Also, applying the substitution $a = x + y$ and $b = x - y$, equation (4) becomes
$$f(a) + f(b) = 2 f\left(\frac{a + b}{2}\right) = f(a + b).$$
This shows $f$ satisfies the Cauchy equation. Thus, (1) can be simplified to
$$f(y f(x)) = x f(y). \tag{5}$$
Putting $y = 1$, we obtain $f(f(x)) = x$. Replacing $x$ by $f(x)$ in (5) and using $f(f(x)) = x$, we get
$$f(x y) = f(x) f(y). \tag{6}$$
It is well-known that a function $f : \mathbb{R} \to \mathbb{R}$ satisfying the Cauchy equation and (6) can only be the zero function or the identity function. Thus, another solution is $f(x) = x$ for all $x \in \mathbb{R}$.