jmc

By AM-GM, $$\begin{array}{rl}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}& =\frac{1}{3x}+\frac{1}{3x}+\frac{1}{3x}+\frac{1}{2y}+\frac{1}{2y}+\frac{1}{z}\\ & \ge 6\sqrt[6]{\frac{1}{3x}\cdot \frac{1}{3x}\cdot \frac{1}{3x}\cdot \frac{1}{2y}\cdot \frac{1}{2y}\cdot \frac{1}{z}}\\ & =6\sqrt[6]{\frac{1}{108x^3{y}^{2}z}}.\end{array}$$Since $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=6,$ this gives us $$x^3{y}^{2}z\ge \frac{1}{108}.$$Equality occurs when $3x=2y=z.$ Along with the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=6,$ we can solve to get $x=\frac{1}{3},$ $y=\frac{1}{2},$ and $z=1,$ so the minimum value is $\boxed{\frac{1}{108}}.$