, 2023

Put $x = 0, y = -1$: $0 = -2f(0) - f(0)$ thus $f(0) = 0$.

Put $y = 1$: $2f(x) = f(2x)$, for all $x$ so the given condition can rewrite as
$$(y + 1)f(x)f(y - 1) = yf(xy) - f(x), \forall x, y$$
Continue to put $y = 0$ then $f(x)f(-1) = -f(x)$. If $f(-1) \neq -1$ then $f(x) = 0$ for all $x$, which is satisfied the given condition. Now suppose that there exist some $x_0$ such that $f(x_0) \neq 0$, this implies that $f(-1) = -1$ and $f(1) = 1$.

In (*), put $x = 1$ then $(y + 1)f(y - 1) = yf(y) - 1$ for all $y$. Replace to the LHS of (*), one can get
$$(yf(y) - 1)f(x) = yf(xy) - f(x)$$
so $f(xy) = f(x)f(y)$, $\forall x, y \neq 0$ which implies that $f$ is multiplication function. Also put $y = -1$ in (*) then $f(x) = -f(-x)$ which implies that $f(x)$ is odd.

Suppose that $a$ is some number such that $f(a) = 0$ then by substituting $x = a$ into (*), we get $0 = yf(ay) - 0$ so $f(ay) = 0, \forall y \neq 0$. If $a \neq 0$ then $ay$ can take any values on $\mathbb{R}$ so $f(x) \equiv 0$, contradiction. Thus $a = 0$ is the unique value such that $f(a) = 0$. Since $(y+1)f(y-1) = yf(y) - 1$, change $y \to -y$ and using the property of odd function, one can get
$$(-y + 1)f(-y - 1) = yf(y) - 1, \forall y \in \mathbb{R}.$$
Thus for all $y \neq \pm 1$ we have
$$(y + 1)f(y - 1) = (y - 1)f(y + 1) \Leftrightarrow \frac{f(y + 1)}{f(y - 1)} = \frac{y + 1}{y - 1}$$
for all $y \neq \pm 1$. Due to the multiplication property of $f$, rewrite the last equation as
$$f\left(\frac{y+1}{y-1}\right) = \frac{y+1}{y-1}, \forall y \in \mathbb{R} \setminus \{\pm 1\}.$$
Note that $t = \frac{y+1}{y-1}$ can take any value on $\mathbb{R} \setminus \{1\}$ so $f(t) = t, \forall t \neq 0, 1$. But $f(1) = 1, f(0) = 0$ so we have $f(x) = x, \forall x \in \mathbb{R}$. Hence, there are two functions that satisfying the given condition: $f(x) = 0, \forall x \in \mathbb{R}$ and $f(x) = x, \forall x \in \mathbb{R}$.