jmc

We can write the product as $$\begin{array}{rl}(2^{1/3})(4^{1/9})(8^{1/27})(16^{1/81})\cdots & =2^{1/3}\cdot (2^2{)}^{1/9}\cdot (2^3{)}^{1/27}\cdot (2^4{)}^{1/81}\cdots \\ & =2^{1/3}\cdot 2^{2/3^2}\cdot 2^{3/3^3}\cdot 2^{4/3^4}\cdots \\ & =2^{1/3+2/3^2+3/3^3+4/3^4+\cdots }.\end{array}$$Let $$S=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\cdots .$$Then $$3S=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\cdots .$$Subtracting these equations, we get $$2S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots =\frac{1}{1-1/3}=\frac{3}{2},$$so $S=\frac{3}{4}.$

Therefore, the infinite product is $2^{3/4}=\boxed{\sqrt[4]{8}}.$