First, setting $y = 0$ to give $f(f(x)^2) = x f(x)$.

Then setting $x = 0$ to get $f(f(y)f(-y) + y^2) = 24y f(0)$. Now change $y$ by $-y$ then the LHS does not change, so we will get
$$24y f(0) = -24y f(0), \forall y \implies f(0) = 0.$$
From $f(f(y)f(-y) + y^2) = 0$, one can get $f(y)f(-y) = -y^2$.

Setting $y = -x$ to get $f(-24x^2) = -24x f(x)$, change $x$ to $-x$ will implies that $f(x)$ is odd. Thus $f(y)^2 = y^2$. We have two solutions: $f(x) = x$ for all $x$ and $f(x) = -x$ for all $x$.

Now we need to handle the point-wise trap. Let $A = \{a \in \mathbb{R} \setminus \{0\} \mid f(a) = a\}$ and $B = \{a \in \mathbb{R} \setminus \{0\} \mid f(b) = -b\}$, suppose that these sets are non-empty. Take $a \in A$ and $b \in B$.

Let $x = a, y = b - a$ to give $f(x + y) = f(b) = -b$ and $f(x - y) = f(2a - b)$. Thus
$$\text{LHS} = f(-b f(2a - b) + 2a(b - a) + (b - a)^2) = \pm(-b \pm (2a - b) + b^2 - a^2).$$
$$\text{RHS} = (b - a)a + a(-b) = -a^2.$$
Thus $\pm(\pm(2a - b)(-b) + b^2 - a^2) = -a^2$. For each pair of $a, b$, there are 4 ways to choose the sign so for some fixed value $a$, there are at most 8 ways to choose $b$ which implies that $|B| \le 8$. Similarly, $|A| \le 8$, this is contradiction since $A \cup B = \mathbb{R} \setminus \{0\}$.

Hence, $f(x) = x$ for all $x$ and $f(x) = -x$ for all $x$. $\square$