Plugging $a = b = 0$ yields $f^2(0) \leq 0$, hence $f(0) = 0$. Again, plugging $b = 0$, and then $a = 0$, we obtain $f(a^2) \leq af(a)$, $\forall a \in \mathbb{R}$ and $f(b^2) \geq bf(b)$, $\forall b \in \mathbb{R}$, therefore $f(x^2) = xf(x)$, $\forall x \in \mathbb{R}$. Replacing the latter in the given equation, we get $f(a)f(b) \leq ab$, $\forall a, b \in \mathbb{R}$.
Also, $-xf(-x) = f((-x)^2) = f(x^2) = xf(x)$, hence $f$ is an odd function.
The last two relations give $f(a)f(b) = -f(a)f(-b) \geq -(-ab) = ab$, so $f(a)f(b) = ab$, $\forall a, b \in \mathbb{R}$.
Thus, $f^2(1) = 1$, which implies $f(1) = \pm 1$ and either $f(x) = x, \forall x \in \mathbb{R}$, or $f(x) = -x, \forall x \in \mathbb{R}$.

It is easy to check that both functions are solutions of the problem.