jmc

By AM-GM, $$x^3+2y^3+4z^3\ge 3\sqrt[3]{x^3\cdot 2y^3\cdot 4z^3}=6xyz.$$Therefore, $$\frac{x^3+2y^3+4z^3}{xyz}\ge 6.$$Equality occurs when $x^3=2y^3=4z^3$; for example, $x=\sqrt[3]{4},$ $y=\sqrt[3]{2},$ and $z=1$ will work, so the minimum value is $\boxed{6}.$