Binomial stuff

Since we need the term that doesn't contain x we could know that we are looking for a term such that $(\sqrt[4]{x})^{6-k} (\frac{1}{\sqrt{x}})^{k}=1$ $(x^{\frac{6-k}{4}})(\frac{1}{x^{\frac{k}{2}}})=\frac{x^{\frac{6-k}{4}}}{x^{\frac{k}{2}}}=1$ $x^{\frac{6-k}{4}-\frac{k}{2}}=1$ we know that every x with exponent equal to zero is 1, so $x^{\frac{6-k}{4}-\frac{k}{2}}=x^{0}$ then $\frac{6-k}{4}-\frac{k}{2}=0$ $\frac{6-k}{4}=\frac{k}{2}$ $6-k=2k$ $6=3k$ $k=\frac{6}{3}=2$ and we know by binomial theorem that the term is $\binom{6}{2}(\sqrt[4]{x})^{6-2} (\frac{1}{\sqrt{x}})^{2})=\boxed{15(\sqrt[4]{x})^{4} (\frac{1}{\sqrt{x}})^{2}}$