$2^{\frac 12}$, $3^{\frac 13}$,$6^{\frac 16}$ put it in an increasing order

Question

$2^{\frac 12}$, $3^{\frac 13}$,$6^{\frac 16}$ put it in an increasing order.

How to solve it without using any calculator. I was trying to understand the behavior of $f(x)=x^{\frac 1x}=e^{\frac {\log x}{x}}$. Then $f'(x)=x^{\frac 1x} \frac{1-\log x}{x^2}$. So it would be zero only when $x=e \in (2,3)$. So it clear that $6^{\frac 16}<3^{\frac 13}$. Now I am confused on how to compare $2^{\frac 12}$, $3^{\frac 13}$ and where will $6^{\frac 16}$ fit at last?

Post comment edit If we raise them to the $6$th power we will get $8,9,6$. Now $6<8<9$ hence $6^{\frac 16}<2^{\frac 12}<3^{\frac 13}$. Thanks for the help @saulspatz in the comment.

Answer 1

Also, $$\sqrt[3]3=\sqrt[6]9$$ and $$\sqrt2=\sqrt[6]8$$ and use that $f(x)=\sqrt[6]x$ increases.