By Holder's Inequality, $\left( x^3 + \frac{y^3}{4} + \frac{z^3}{9} \right) (1 + 2 + 3)(1 + 2 + 3) \ge (x + y + z)^3 = 12^3,$ so $x^3 + \frac{y^3}{4} + \frac{z^3}{9} \ge \frac{12^3}{6^2} = 48.$ Equality occurs when $(x,y,z) = (2,4,6)$ .

Also, $x^3 + \frac{y^3}{4} + \frac{z^3}{9} \le (x + y + z)^3 = 1728.$ Equality occurs when $(x,y,z) = (12,0,0)$ . Thus, the minimum is 48 and the maximum is 1728.

Since the function is convex, we can use Jensen's inequality:

$f(x,y,z)=x^3+2\left(\frac{y}{2}\right)^3+3\left(\frac{z}{3}\right)^3\geq 6\left(\frac{x+y+z}{6}\right)^3=6*2^3=48=f(2,4,6)=m$

Again, since the function is convex, the maximum $M$ must be attained at one of the vertices; checking the three cases we see that $M=f(12,0,0)=1728$ .

Thus the required sum is $M+m=\boxed{1776}$

Disclaimer: This solution is currently wrong.

I'm going to find the minimum value via Applying the Arithmetic Mean Geometric Mean Inequality and the maximum value via differentiation followed by Second derivative test.

We are given that $x+ y+z =12$ . Apply AM-GM on the numbers $\{ x^3, x^3 , x^3 ,8,8,8,8,8,8 \}$ , we get $\dfrac{3x^3 + 6\cdot 2^3}9 \ge \sqrt[9]{x^9 \cdot 2^{18}} \Rightarrow x^3 + 16 \ge 12x \quad\quad\quad\quad\quad (1)$ .

Similiarly, apply AM-GM on the numbers $\{y^3,y^3,y^3,64,64,64,64,64,64\}$ , we get $\dfrac{y^3+128}4 \ge 12y \quad\quad\quad\quad\quad (2)$ .

And again, for the numbers $\{z^3 , 216,216\}$ , we get $\dfrac{z^3+432}9 \ge 12z \quad\quad\quad\quad\quad (3)$ .

Adding these 3 inequalities and apply the fact that $x+y+z=12$ , we get $x^3 + \frac{y^3}4 + \frac{z^3}9 \ge 48$ .

Equality holds when $x=2,y=4,z=6$ . Thus the minimum value is 48.

Now for the maximum value, we must check on the boundary values:
Case 1 : $x = 0$ only.
Case 2 : $y= 0$ only.
Case 3 : $z= 0$ only.
Case 4 : $x =y= 0$ only.
Case 5 : $x =z= 0$ only.
Case 6 : $y =z= 0$ only.

It's easy to check that for Case 4, 5, 6, the value of the expression is question are 192, 432, 1728 respectively.

For Case 1 , the constraint simplifies to $y+z=12$ , and we want to maximize $\frac{y^3}4 + \frac{z^3}9$ . Apply the substitution of $z = 12 - y$ onto the expression $\frac{y^3}4 + \frac{z^3}9$ and differentiate it shows that its maximum value is $\frac{1728}{25}$ . And we can prove that it's a maximum value via Second Derivative Test.

Analogously, we can solve for Case 2 and 3, which yield the maximum value of 108 and 192 respectively.

Hence from all these 6 cases, the maximum value of the expression in question is 1728 at $y=z=0$ , $x = 12$ .