If $y=kx^{\frac{1}{4}}$ and $y=3\sqrt{2}$ at $x=81$, what is the value of $y$ at $x=4$?
First we have to solve for $k$ where $3\sqrt{2}=k\cdot81^{\frac{1}{4}}$.  Since $81^{\frac{1}{4}}=3$, we have $3\sqrt{2}=k\cdot3$, so $k = \sqrt{2}$.

When $x=4$, we have$$y=k\cdot4^{\frac{1}{4}}=k\cdot\sqrt{2}.$$Since $k=\sqrt{2}$, we have $$y=\sqrt{2}\cdot\sqrt{2}=\boxed{2}.$$