Since the inequality is cyclic, w.l.o.g. let $\min\{a, b, c\} = c$. Then $a = c + x$, $b = c + y$ for some nonnegative $x$ and $y$. After substitution $a = c + x$ and $b = c + y$ we get
$$\begin{align*} a^3 + b^3 + c^3 - 3abc &= (c + x)^3 + (c + y)^3 + c^3 - 3(c + x)(c + y)c \\ &= (3c + x + y)(x^2 - xy + y^2) \\ ab^2 + bc^2 + ca^2 - 3abc &= (c + x)(c + y)^2 + (c + y)c^2 + c(c + x)^2 - 3(c + x)(c + y)c \\ &= (x^2 - xy + y^2)c + x y^2 \end{align*}$$
Therefore, in order to prove $(*)$ we have to show that for each $c > 0$ and $x, y \ge 0$
$$(3 - M)(x^2 - xy + y^2)c + x^3 + y^3 - M x y^2 \ge 0 \quad (1)$$
For $x = 1$, $y = \sqrt[3]{2}$ and any $c > 0$
$$(3 - M)(1 - \sqrt[3]{2} + \sqrt[3]{4})c + 3 - \sqrt[3]{4}M \ge 0 \quad (2)$$
Now let us show that $3 \ge \sqrt[3]{4}M$. Suppose that $3 < \sqrt[3]{4}M$. If $3 - M \le 0$ then (1) is not held for any $c > 0$. If $3 - M > 0$ then for
$$c < \frac{\sqrt[3]{4}M - 3}{(3 - M)(1 - \sqrt[3]{2} + \sqrt[3]{4})}$$
(2) is not held. Thus, $3 - \sqrt[3]{4}M \ge 0$. Now let us show that for $M = \frac{3}{\sqrt[3]{4}}$ the inequality (1) holds. Since $M < 3$, $(3 - M)(x^2 - xy + y^2)c \ge 0$ and the required result will follow from $x^3 + y^3 \ge M x y^2$ which in turn is a consequence of AM-GM inequality:
$$x^3 + y^3 = x^3 + \frac{y^3}{2} + \frac{y^3}{2} \ge \frac{3}{\sqrt[3]{4}} x y^2 = M x y^2$$
Thus, the greatest $M$ is $\frac{3}{\sqrt[3]{4}}$.