Let $P(x, y)$ denote the given functional equation.

$$P(x, 0) \rightarrow f(xf(0) - f(x)) = -x - f(0).$$

From this we know that $f$ is surjective, and $f(f(x)) = x$. Note that
$$P(1, 1) \rightarrow f(-1) = -1 \text{ and } P(0, -1) \rightarrow f(1 - f(0)) = 1 - f(0).$$

Consider $x = 0, y = 1 - f(0)$:
$$-1 = f(-1) = (1-f(0))f(0) - f(1-f(0)) = -(1-f(0))^2 \Leftrightarrow f(0) = 0 \text{ or } 2.$$

If $f(0) = 2$, then $P(x, 0)$ becomes
$$f(2x - f(x)) = -x - 2 = f(f(-x - 2))$$
By injectivity we obtain the following system of equations
$$\begin{cases} 2x - f(x) = f(-x - 2) \\ 2(-x - 2) - f(-x - 2) = f(x) \end{cases}$$
which is clearly a contradiction. Therefore $f(0) = 0$ and $f(-x) = -f(x)$. In particular, $f(1) = 1$. Since $f$ is an odd function,

$$P(x, f(-x)) \rightarrow f(-x^2) = -f(x)^2 \Leftrightarrow f(x)^2 = f(x^2).$$

Finally, according to the following identity,
$$P(x, 1) \rightarrow f(x - f(x) - 1) = f(x) - x - 1.$$
Suppose $x - f(x) = \alpha$, and recursing on the above identity, by induction we obtain
$$f(\alpha 2^n - 1) = -\alpha 2^n - 1 \text{ holds for all } n \in \{0\} \cup \{\text{positive integers}\}.$$

Clearly $\alpha \neq 0$ would contradict $f(x^2) = f(x)^2 \geq 0$, therefore
$$f(x) = x, \forall x \in \mathbb{R}.$$