jmc

By the Binomial Theorem, the coefficient of $a^4{b}^2$ in the expansion of $(a+b{)}^6$ is $\left(\genfrac{}{}{0ex}{}{6}{4}\right).$ Now, $$x^2=\left(x^4\right)\left({\left(\frac{1}{x}\right)}^2\right).$$ Thus the coefficient of of $x^2$ is the coefficient of $x^4(1/x{)}^2,$ namely $\left(\genfrac{}{}{0ex}{}{6}{4}\right)=\boxed{15}.$