10. Try to transform the left side of the given inequality into a form that can apply the mean value inequality, and note the connection between this expression and $\left(a^{2}+b^{2}+c^{2}\right)^{2}$, thereby finding the minimum value of $M$.
First consider $P(t)=t b\left(t^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c t\left(c^{2}-t^{2}\right)$
It is easy to see that $P(b)=P(c)=P(-c-b)=0$.
Then,
$$|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)|$$
$$=|P(a)|=|(b-c)(a-b)(a-c)(a+b+c)|.$$

Thus, the original inequality is equivalent to:
$$|(b-c)(a-b)(a-c)(a+b+c)| \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2}.$$

By symmetry, without loss of generality, assume $a \leqslant b \leqslant c$, then
$$|(b-c)(a-b)|=(b-a)(c-b) \leqslant\left[\frac{(b-a)+(c-b)}{2}\right]^{2}=\frac{(c-a)^{2}}{4},$$

The equality holds if and only if $b-a=c-b$, i.e., $2 b=a+c$.
Note that $\left[\frac{(b-a)+(c-b)}{2}\right]^{2} \leqslant \frac{(c-b)^{2}+(b-a)^{2}}{2}$
$$\begin{array}{l} \Leftrightarrow 3(c-a)^{2} \\ \leqslant 2\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right], \end{array}$$
The equality holds if and only if $2 b=a+c$. Therefore,
$$\begin{array}{l} |(b-c)(a-b)(a-c)(a+b+c)| \\ \leqslant \frac{1}{4}\left|(c-a)^{3}(a+b+c)\right|=\frac{1}{4} \sqrt{(c-a)^{6}(a+b+c)^{2}} \\ \leqslant \frac{1}{4} \sqrt{\left\{\frac{2\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right]}{3}\right\}^{3}(a+b+c)^{2}} \\ \leqslant \frac{\sqrt{2}}{2}\left\{\sqrt[4]{\left[\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}\right]^{3}(a+b+c)^{2}}\right\}^{2}. \end{array}$$

By the mean value inequality,
$$\begin{array}{l} |(b-c)(a-b)(a-c)(a+b+c)| \\ \leqslant \frac{\sqrt{2}}{2}\left[\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}+(a+b+c)^{2}}{4}\right]^{2} \\ =\frac{9 \sqrt{2}}{32}\left(a^{2}+b^{2}+c^{2}\right)^{2}. \end{array}$$

Thus, $M \geqslant \frac{9 \sqrt{2}}{32}$, and the equality holds if and only if $2 b=a+c$
and $\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}=(a+b+c)^{2}$.
Solving, we get $2 b=a+c, (c-a)^{2}=18 b^{2}$.
Taking $b=1$, we get $a=1-\frac{3 \sqrt{2}}{2}, c=1+\frac{3 \sqrt{2}}{2}$. At this point, the equality in the original inequality holds. Therefore, $M_{\text {min }}=\frac{9 \sqrt{2}}{32}$.