We use the inequality of arithmetic–geometric mean as follows:

$$\sqrt[3]{a^2+2bc} = \frac{1}{\sqrt[3]{12^2}} \sqrt[3]{(a^2+2bc) \cdot 12 \cdot 12} \le \frac{1}{\sqrt[3]{12^2}} \cdot \frac{a^2+2bc+12+12}{3} = \frac{1}{3\sqrt[3]{12^2}} (a^2+2bc+24),$$
$$\sqrt[3]{b^2+2ca} = \frac{1}{\sqrt[3]{12^2}} \sqrt[3]{(b^2+2ca) \cdot 12 \cdot 12} \le \frac{1}{\sqrt[3]{12^2}} \cdot \frac{b^2+2ca+12+12}{3} = \frac{1}{3\sqrt[3]{12^2}} (b^2+2ca+24),$$
$$\sqrt[3]{c^2+2ab} = \frac{1}{\sqrt[3]{12^2}} \sqrt[3]{(c^2+2ab) \cdot 12 \cdot 12} \le \frac{1}{\sqrt[3]{12^2}} \cdot \frac{c^2+2ab+12+12}{3} = \frac{1}{3\sqrt[3]{12^2}} (c^2+2ab+24),$$
from which we get
$$\begin{aligned} S &= \sqrt[3]{a^2+2bc} + \sqrt[3]{b^2+2ca} + \sqrt[3]{c^2+2ab} \\ &\le \frac{1}{3\sqrt[3]{12^2}} (a^2+b^2+c^2+2ab+2bc+2ca+72) \\ &= \frac{1}{3\sqrt[3]{12^2}} \left[ (a+b+c)^2 + 72 \right] = \frac{36}{\sqrt[3]{12^2}} = \frac{18}{\sqrt[3]{18}} = 3\sqrt[3]{12}. \end{aligned}$$
Equality holds when
$$a^2 + 2bc = 12, \quad b^2 + 2ca = 12, \quad c^2 + 2ab = 12$$
$$\Leftrightarrow (a-b)(a+b-2c) = 0, \quad (b-c)(b+c-2a) = 0, \quad c^2 + 2ab = 12$$
$$\Leftrightarrow (a-b)(6-3c) = 0, \quad (b-c)(6-3a) = 0, \quad c^2 + 2ab = 12$$
$$\Leftrightarrow a = b = c = 2.$$
Therefore the maximal value of the expression is $3\sqrt[3]{12}$ and it is obtained for $a = b = c = 2$.

Solution 2:
Solution with Hölder's inequality (P. Lolas, G. Vlachos, M. Agelis)

We have
$$\begin{aligned} \sqrt[3]{9(a+b+c)^2} &= \sqrt[3]{9 \cdot \sqrt[3]{(a^2+2bc)+(b^2+2ca)+(c^2+2ab)}} \\ &= \sqrt[3]{1+1+1} \cdot \sqrt[3]{(a^2+2bc)+(b^2+2ca)+(c^2+2ab)} \\ &\ge \sqrt[3]{a^2+2bc} + \sqrt[3]{b^2+2ca} + \sqrt[3]{c^2+2ab} = S. \end{aligned}$$
Hence we have $S \le \sqrt[3]{9 \cdot 6^2} = 3\sqrt[3]{12}$, where equality holds for
$$a^2 + 2bc = b^2 + 2ca = c^2 + 2ab, \quad a+b+c=6$$
$$\Leftrightarrow (a-b)(a+b-2c) = 0, \quad (b-c)(b+c-2a) = 0, \quad a+b+c = 6$$
$$\Leftrightarrow (a-b)(6-3c) = 0, \quad (b-c)(6-3a) = 0, \quad a+b+c = 6$$
$$\Leftrightarrow a=b=c=2.$$

Solution 3:
Solution using Cauchy–Schwarz inequality (X. Tsampasidis)

We put $x = \sqrt[3]{a^2 + 2bc}$, $y = \sqrt[3]{b^2 + 2ca}$, $z = \sqrt[3]{c^2 + 2ab}$. Then $x, y, z > 0$ and from Cauchy–Schwarz inequality we get
$$\begin{aligned} & (1^3 + 1^3 + 1^3)(1^3 + 1^3 + 1^3)(x^3 + y^3 + z^3) \ge (1 \cdot 1 \cdot x + 1 \cdot 1 \cdot y + 1 \cdot 1 \cdot z)^3 \\ & \Leftrightarrow 9(x^3 + y^3 + z^3) \ge (x+y+z)^3. \end{aligned}$$
However we have
$$x^3 + y^3 + z^3 = a^2 + 2bc + b^2 + 2ca + c^2 + 2ab = (a+b+c)^2 = 36.$$
Thus the last inequality becomes
$$x + y + z \le \sqrt[3]{9(x^3 + y^3 + z^3)} = \sqrt[3]{9 \cdot 36} = 3\sqrt[3]{12}.$$
Equality holds when $x = y = z, a + b + c = 6$ or equivalently
$$a^2 + 2bc = b^2 + 2ca = c^2 + 2ab, \quad a + b + c = 6 \Leftrightarrow a = b = c = 2.$$

Solution 4:
Solution using Jensen's inequality (K. Axiotis, N Athanasiou, D. Sotiriou)

The function $f(x) = \sqrt[3]{x}, x > 0$, is twice differentiable in $(0, +\infty)$, with $f'(x) = \frac{1}{3}x^{-\frac{2}{3}}$ and $f''(x) = -\frac{2}{9}x^{-\frac{5}{3}}$. Hence $f(x) = \sqrt[3]{x}$ is concave in $(0, +\infty)$. Hence from Jensen's inequality we get
$$\begin{aligned} & \frac{1}{3}\left[f(a^2+2bc)+f(b^2+2ca)+f(c^2+2ab)\right] \le f\left(\frac{a^2+2bc+b^2+2ca+c^2+2ab}{3}\right) \\ \Rightarrow & f(a^2+2bc)+f(b^2+2ca)+f(c^2+2ab) \le 3f\left(\frac{(a+b+c)^2}{3}\right) = 3f(12) = 3\sqrt[3]{12}. \\ \Rightarrow & \sqrt[3]{a^2+2bc} + \sqrt[3]{b^2+2ca} + \sqrt[3]{c^2+2ab} \le 3\sqrt[3]{12}. \end{aligned}$$
Since for $a=b=c=2$ the first part takes the value $3\sqrt[3]{12}$, it follows that the maximal value of the expression $S$ is $3\sqrt[3]{12}$.

Solution 5:
Solution using power means inequality (A. Mousatov)

We put $x = \sqrt{a^2 + 2bc}$, $y = \sqrt{b^2 + 2ca}$, $z = \sqrt{c^2 + 2ab}$. Then we have
$$x + y + z = (a + b + c)^2 = 36.$$
Hence, if we put $u = \sqrt[3]{x}$, $v = \sqrt[3]{y}$, $w = \sqrt[3]{z}$, then we get $S = u + v + w$ and from power means inequality we obtain
$$S = u + v + w \le 3 \cdot \sqrt[3]{\frac{u^3 + v^3 + w^3}{3}} = 3 \cdot \sqrt[3]{12},$$
since $u^3 + v^3 + w^3 = x + y + z = 36$. Equality is valid when
$$u = v = w \Leftrightarrow a = b = c = 2.$$

Solution 6:
Solution using rearrangement inequality (K. Dermentzis)

We put $x = \sqrt[3]{a^2 + 2bc}$, $y = \sqrt[3]{b^2 + 2ca}$, $z = \sqrt[3]{c^2 + 2ab}$. Then we have
$$S = x + y + z \text{ and } x^3 + y^3 + z^3 = (a+b+c)^2 = 36.$$
Next we put
$$S_1 = x^3 + y^3 + z^3 = x^2x + y^2y + z^2z, \quad S_2 = x^2y + y^2z + z^2x, \quad S_3 = x^2z + y^2x + z^2y$$
And since the triads $(x^2, y^2, z^2)$, $(x, y, z)$ have the same ordering, from rearrangement inequality we obtain $S_1 \ge S_2$ and $S_1 \ge S_3$. Moreover, since $x, y, z > 0$, we have $S_1 = x^3 + y^3 + z^3 \ge 3xyz$, and therefore
$$\begin{aligned} S^3 &= x^3 + y^3 + z^3 + 3(x^2y + y^2z + z^2x) + 3(x^2z + y^2x + z^2y) + 6xyz \\ &= S_1 + 3S_2 + 3S_3 + 6xyz \le S_1 + 3S_1 + 3S_1 + 2S_1 = 9S_1. \end{aligned}$$
Hence
$$S = x + y + z \le \sqrt[3]{9(x^3 + y^3 + z^3)} = \sqrt[3]{9 \cdot 12} = 3 \cdot \sqrt[3]{12}.$$
Since for $a=b=c=2$ the expression $S$ takes the value $3\sqrt[3]{12}$, it follows that its maximal value is $3\sqrt[3]{12}$.