jmc

First we have to solve for $k$ where $3\sqrt{2}=k\cdot 81^{\frac{1}{4}}$. Since $81^{\frac{1}{4}}=3$, we have $3\sqrt{2}=k\cdot 3$, so $k=\sqrt{2}$.

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