Let $\mathcal{P}(a, b, c)$ denote the expression
$$f(a + b + c)f(ab + bc + ca) = f(a)f(b)f(c) + f(a + b)f(b + c)f(c + a).$$

$\mathcal{P}(0, 0, 0) \Rightarrow f(0)^2 = 2f(0)^3.$ So, $f(0) = \frac{1}{2}$ or $f(0) = 0$. Exclude the solution $f \equiv 0$.

Case 1 ($f(0) = \frac{1}{2}$):
$\mathcal{P}(a, 0, 0) \Rightarrow \frac{1}{4}f(a) = \frac{1}{2}f(a)^2 \Rightarrow \forall a, f(a) = 0$ or $f(a) = \frac{1}{2}$.

Now, if there is $r \neq 0$ such that $f(r) = \frac{1}{2}$, take any $a \in \left(-\frac{|r|}{3}, \frac{|r|}{3}\right)$. Then, since $(r + 3a)(r - a) > 0$, we will find real solutions to $x^2 + (a - r)x + a(a - r) = 0$, which we call $b$ and $c$. Then, $(a + b + c) = r$ and $(ab + bc + ca) = 0$. Therefore, $\mathcal{P}(a, b, c) \Rightarrow f(a)f(b)f(c) + f(a + b)f(b + c)f(c + a) = \frac{1}{4}$, so $f(a) = \frac{1}{2}$. So $f(a) = \frac{1}{2}$ for all $a \in \left(-\frac{|r|}{3}, \frac{|r|}{3}\right)$.

If $f(a+b) = 0$, then $\mathcal{P}(a, b, 0) \Rightarrow \frac{1}{2}f(a)f(b) = 0$, so $f(a) = 0$ or $f(b) = 0$. Therefore, if $f(a) = f(b) = \frac{1}{2}$, then $f(a+b) = \frac{1}{2}$. Since $f = \frac{1}{2}$ in an interval containing $0$, we can therefore conclude $f = \frac{1}{2}$ on $\mathbb{R}$. Clearly, $f \equiv \frac{1}{2}$ is a solution.

We claim that $f(0) = \frac{1}{2}$ and $f(x) = 0$ for all $x \neq 0$ is a solution to the functional equation. If $a = b = c = 0$, or $(a+b) = (b+c) = (c+a) = 0$, then $\mathcal{P}(a, b, c)$ is satisfied. So, we only need to show that if $(a+b+c) = (ab+bc+ca) = 0$, then $a = b = c = 0$. However, in this case, $a, b, c$ are the three roots (with multiplicity) of $x^3 - abc = 0$, and the only way they can all be real is if $a = b = c = 0$.

Case 2 ($f(0) = 0$):
Define sets $R = \{x \in \mathbb{R} \mid f(x) = 0\}$, $S = \{x \in \mathbb{R} \mid f(x) \neq 0\}$. We know $S$ is nonempty and does not contain $0$.

Claim 1. There are arbitrarily large negative numbers in $S$.

Proof. Suppose $c \in S$. Then $\mathcal{P}(c, c, c) \Rightarrow 2c \in S$ or $3c \in S$. This implies that $S$ cannot be bounded on both sides, and if $S$ contains a negative real then it contains arbitrarily large negative reals. So suppose $S$ does not contain any negative reals, i.e. $f(x) = 0$ for negative $x$. Then, for $2b > a > b > 0$,
$$\mathcal{P}(a, -b, a) \Rightarrow f(a - b)^2 f(2a) = f(a - b)f(a^2 - 2ab) + f(a)^2 f(-b) = 0.$$
Let $c \in S$. Take $b > c$ and $a = b + c$. Then $f(2a) = 0$. So, $f(x) = 0$ for all $x > 4c$. So, $S \subset (0, 4c]$. This contradicts that $S$ cannot be bounded on both sides. $\square$

Claim 2. $c, d \in R \Rightarrow (c + d) \in R$.

Proof. If $a + b = d$, then $\mathcal{P}(a, b, c) \Rightarrow f(c + d)f(ab + cd) = 0$.
Varying $a, b$ over all reals such that $a + b = d$, $ab$ takes all values in $(-\infty, \frac{d^2}{4}]$. Since $S$ contains arbitrarily large negative numbers, we can find $f(ab + cd) \neq 0$ for some $a, b$ with $a + b = d$. Therefore $f(c + d) = 0$. $\square$

Claim 3. $r \in R \iff -r \in R$. Also, $s \in S \iff -s \in S$.

Proof. Take $-r \in R$. By Claim 2, $-nr \in R$ for all $n \in \mathbb{N}$.
$$\mathcal{P}(2r, 2r, -r) \Rightarrow f(r)^2 f(4r) = f(3r)f(0) + f(2r)^2 f(-r) = 0$$
So, $r \in R$ or $8r \in R$. If $8r \in R$, then so is $8r + (-7r) = r$. So, $r \in R$ in either case.
Therefore $r \in R \iff -r \in R$. Since $S = \mathbb{R} \setminus R$, the other equivalence follows. $\square$

Claim 4. If $r \in R \setminus \{0\}$ and $s \in S$, then $\frac{s}{r} \in R$.

Proof.
$$\mathcal{P}\left(r, \frac{s}{r}, 0\right) \Rightarrow f\left(r+\frac{s}{r}\right) f(s) = f\left(r+\frac{s}{r}\right) f(r) f\left(\frac{s}{r}\right) = 0 \Rightarrow f\left(r+\frac{s}{r}\right) = 0$$
Using Claim 2 and Claim 3, we get that $\frac{s}{r} = \left(r + \frac{s}{r}\right) + (-r) \in R$.

Claim 5. $r \in R, s \in S \Rightarrow (r+s) \in S, rs \in R$

Proof. If $(r+s) \in R$, we will get $s = (r+s)+(-r) \in R$ , which is a contradiction. Thus, $(r+s) \in S$. Now,
$$\mathcal{P}(r, s, 0) \Rightarrow f(rs) = f(r)f(s) = 0 \Rightarrow rs \in R.$$

Claim 6. $f$ is injective at $0$, i.e. $R = \{0\}$

Proof. We know there is some $s \in S$. Assume that we have $r \in R \setminus \{0\}$. By Claim 5 and Claim 3, $(r+s), (-s) \in S$. Also Claim 4 implies $\frac{s}{r} \in R$. Therefore, by Claim 5, we get $(\frac{s}{r}) \cdot (r+s) \in R$ and $(\frac{s}{r}) \cdot (-s) \in R$. Then, by Claim 2, $s = (\frac{s}{r}) \cdot (r+s) + (\frac{s}{r}) \cdot (-s) \in R$, which is a contradiction.
Therefore, we can conclude that $R = \{0\}$.

So, for any $a, b$ such that $(a+b) \neq 0$, $\mathcal{P}(a,b,0) \Rightarrow f(ab) = f(a)f(b)$.
Given $c \in S$, $\mathcal{P}(a, -a, c) \Rightarrow f(a)f(-a)f(c) = f(-a^2)f(c) \Rightarrow f(-a^2) = f(a)f(-a)$. These two statements combined imply that $f$ is multiplicative. So $f(1) = 1$.

Now, take $abc = q$ and $(a+b)(b+c)(c+a) = p$. Then, since $f$ is multiplicative,
$$f(p+q) = f((a+b+c)(ab+bc+ca)) = f((a+b)(b+c)(c+a)) + f(abc) = f(p) + f(q)$$

Lemma. Given any $p, q \in \mathbb{R} \setminus \{0\}$, we can find $a, b, c \in \mathbb{R}$ such that $(a+b)(b+c)(c+a) = p$ and $abc = q$.

Proof. If we find $a, b, c$ such that $\frac{(a+b)(b+c)(c+a)}{abc} = \frac{p}{q}$, then by scaling $a, b, c$ by $\left(\frac{q}{abc}\right)^{1/3}$, we will get appropriate numbers.
Now if we take $b \neq 0$ and $c \neq 0$ to have opposite signs such that $(b+c) \neq 0$, then $\frac{(a+b)(b+c)(c+a)}{abc} = \frac{p}{q} \iff a^2 + (b+c)a + bc = a \cdot \frac{pbc}{q(b+c)}$. Treating this as a quadratic in $a$, the discriminant is positive. Now, take $a$ to be one of the roots.

Therefore $f$ is additive. So, for any $q \in \mathbb{Q}$, we have $f(q) = qf(1) = q$. However, since $f: \mathbb{R} \to \mathbb{R}$ is multiplicative as well, $f(x) > 0$ for all $x > 0$, so $f$ is monotonically increasing. Since rationals are dense in $\mathbb{R}$, we get $f(x) = x$ for all $x \in \mathbb{R}$.

Final answer:
The solutions are:
- $f(x) \equiv \frac{1}{2}$ for all $x \in \mathbb{R}$.
- $f(0) = \frac{1}{2}$ and $f(x) = 0$ for all $x \neq 0$.
- $f(x) = x$ for all $x \in \mathbb{R}$.