Find the middle term of the expansion of $\left(\sqrt{x} - \dfrac{1}{x}\right)^6$
There are $7$ terms in the expansion of $\left(\sqrt{x} - \dfrac{1}{x}\right)^6$.
The $4^{th}$ term will be the middle term.
Now, the $\left(r + 1\right)^{th}$ term $T_{r+1}$ in the expansion of $\left(p + q\right)^n$ is
$T_{r+1} = C^n_r \; p^{n-r} \; q^r$
$\therefore \;$ In the given problem,
Fourth term $= T_4 = T_{3+1} = C^{6}_{3} \; \left(\sqrt{x}\right)^{6-3} \; \left(\dfrac{-1}{x}\right)^3$
$\begin{aligned}
i.e. \; T_4 & = \dfrac{6!}{3! \left(6-3\right)!} \times \left(\sqrt{x}\right)^3 \times \left(\dfrac{-1}{x^3}\right) \\\\
& = \dfrac{-6 \times 5 \times 4 \times 3!}{3! \times 6} \times x^{\frac{3}{2} - 3} \\\\
& = -20 x^{\frac{-3}{2}}
\end{aligned}$