Setting $x = 0$ yields $f(0) = f(0)f(y) + yf(0)$. If $f(0) \neq 0$, we obtain $1 = f(y) + y$ or equivalently $f(y) = 1 - y$ for all $y \in \mathbb{R}$. Inserting this in the original equation yields
$$1 - x^2 - xy = (1-x)(1-y) + y(1-x) + x(1-x-y)$$
for $x, y \in \mathbb{R}$, which is true.
Therefore, we are left with $f(0) = 0$.
With the substitution $x + y = z$, the functional equation is equivalent to
$$f(xz) = f(x)f(z-x) + (z-x)f(x) + xf(z). \quad (1)$$
Exchanging $x$ and $z$ yields
$$f(zx) = f(z)f(x-z) + (x-z)f(z) + zf(x). \quad (2)$$
Combining (1) and (2) yields
$$f(x)f(z-x) + (z-x)f(x) + xf(z) = f(z)f(x-z) + (x-z)f(z) + zf(x)$$
or equivalently
$$f(x)(f(z-x) - x) = f(z)(f(x-z) - z). \quad (3)$$
Setting $z = 0$ yields
$$f(x)(f(-x) - x) = f(0)f(x) = 0,$$
so for each $x \in \mathbb{R}$ we either have $f(x) = 0$ or $f(-x) = x$.
Assume that there is an $x \neq 0$ with $f(x) = 0$. For $z \neq 0$, we have $f(x-z) \in \{0, z-x\}$, so $f(x-z) \neq z$. Thus (3) implies $f(z) = 0$ for all $z \neq 0$. It is clear that the constant function $f = 0$ is a solution.
Otherwise, we have $f(x) \neq 0$ for all $x \neq 0$ and therefore $f(x) = -x$ for all $x$. This is also a solution.
We conclude that there are three solutions $f$, namely $f = 0$, $f(x) = -x$ and $f(x) = 1 - x$.

Setting $x = 0$ gives $f(0) = f(0)f(y) + yf(0)$. Since $f(y) = 1 - y$ gives a solution, we remain with the case $f(0) = 0$. Setting $x = 1$ gives
$$f(1+y) = f(1)f(y) + yf(1) + f(1+y), \quad \text{i. e.} \quad 0 = f(1)(f(y) + y).$$
Since $f(y) = -y$ is a solution, we remain with the case $f(1) = 0$.

Now suppose $f(0) = f(1) = 0$. Setting $y = -x$ gives $0 = f(x)f(-x) - xf(x)$, that is
$$f(x) = 0 \quad \lor \quad f(-x) = x. \tag{4}$$
Setting $y = 1 - x$ gives $f(x) = f(x)f(1-x) + (1-x)f(x)$, that is
$$f(x) = 0 \quad \lor \quad f(1-x) = x. \tag{5}$$
If $f(x) \neq 0$ for some $x \neq 0$, we have $f(-x) = x \neq 0$ and $f(1-x) = x \neq 0$ and applying (4) and (5) to $-x$ and $1-x$ instead of $x$ we get $f(x) = -x = 1-x$, a contradiction. Hence $f(x) = 0$ for all $x \in \mathbb{R}$.