The answers are $f(x) = x$, $f(x) = -x$ and $f(x) = 0$.

First, let's prove that $f(x)$ is injective at point $0$. Assume there exist two distinct real numbers $t_1$ and $t_2$ such that $f(t_1) = f(t_2) = 0$. Comparing $P(-f(b) - f(0), b, t_1)$ and $P(-f(b) - f(0), b, t_2)$ gives us
$$(f(b) + f(0)) b t_1 = (f(b) + f(0)) b t_2.$$
So if $f(x)$ is not injective at point $0$, we have
$$\Rightarrow \forall b \neq 0 : f(b) = -f(0).$$
Which gives us $f(x) = 0$ as an answer.

So if $f(x)$ is a non-constant function, then it must be injective at point $0$.

Now, $P(-f(0) - f(f(0)), 0, 0)$ gives us $f(-f(0) - f(f(0))) = 0$ and we have a real number $t$ such that $f(t) = 0$. $P(-f(f(0)), t, 0)$ gives us
$$f(-f(f(0))) = f(-f(0) - f(f(0))) = 0.$$
And according to the injectivity at point $0$, it follows that $f(0) = 0$.

Now, $P(-f(f(b)), b, 0)$ and $P(-f(f(c)), 0, c)$ give us
$$f(b) f(f(b))^2 = b^2 f(b).$$
If $b \neq 0$ then $f(b) \neq 0$ therefore
$$f(f(b)) = \pm b \quad \forall b \neq 0.$$
And since $f(f(0)) = f(0) = 0$, we have
$$f(f(b)) = \pm b \quad \forall b \in \mathbb{R}.$$
If there exists a real number $c \neq 0$ such that $f(f(c)) = -c$, $P(c, 0, c)$ gives us
$$f(c)(f(c)^2 + c^2) = 0$$
which is a contradiction.

So
$$\forall c \in \mathbb{R}: f(f(c)) = c$$
Now $P(-a, 0, a)$ gives us $f(-a)^2 + a^2 f(a) = 0$ and $P(a, 0, -a)$ gives us $f(a)^3 + a^2 f(-a) = 0$ and they lead to $f(a) = \pm a$.

If there exist non-zero real numbers $b, c$ such that $f(b) = b$ and $f(c) = -c$, $P(-b - c, b, c)$ leads to contradiction. So $f(x) = x$ and $f(x) = -x$ are the only solutions.

$f(x) = 0$ is also a solution.

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