jmc

When subtracting, we notice that we can't perform the subtraction of the right-most digits unless we borrow. However, we cannot borrow from the $6$'s digit, since it is $0$, so we must borrow from the $36$'s digit. Borrowing allows us to replace $1\cdot 36^2+0\cdot 6+1$ with $0\cdot 36+6\cdot 6+1$, which in turn we can borrow to replace with $0\cdot 36+5\cdot 6+7$. Now, we can subtract directly to find that: \begin{array}{c@{}c@{\;}c@{\ }c@{\ }c@{\ }c} & & & \cancelto{0}{1} & \cancelto{5}{0} & \cancelto{7}{1}_{6} \\ &- & & & 3 & 2_{6} \\ \cline{2-6} && & & 2 & 5_{6} \\ \end{array}Thus, the answer is $\boxed{25_6}$.