Let $X_1,\ldots,X_n$ be iid. $N(\mu,\sigma^2)$. Let $T=(\bar{X},S^2)$, sample mean and sample variance, be a sufficient statistic. I'd like to show that $$U=\frac{X_{(n)}-\bar{X}}{X_{(n)}-X_{(1)}}$$
is independent of $T.$ $(1)$ indicates the smallest value and $(n)$ the biggest. Basu's theorem gives us that if $U$ is ancillary, we have independence. So I would like to show that $U$ is ancillary, ie. that is does not depend on $\mu$ or $\sigma$. I was given the hint to express $U$ in terms of $Z_i=(X_i-\mu)/\sigma$ and I intend on using it, but I couldn't get there.
Edited: I've included the wrong $U$, it's fixed.
Let's be clear about a couple of things.
You wrote: "is ancillary, ie. that is does not depend on $μ$ or $σ$."
The definition should say that its distribution does not depend on $\mu$ or $\sigma.$
It should also say that it's a random variable that is observable, i.e. it is a statistic. For example, $(X_1-\mu)/\sigma$ is a random variable whose distribution does not depend on $\mu$ or $\sigma,$ but it is not an ancillary statistic since it is not a statistic. Your $U$ is clearly observable, so what is left is to show that the distribution of $U$ does not depend on the pair $(\mu,\sigma).$
By some algebra, canceling $\mu$s and $\sigma$s where possible, you can show that $$ \frac{X_{(n)}-\bar{X}}{X_{(n)}-X_{(1)}} = \frac{Z_{(n)}-\bar{Z}}{Z_{(n)}-Z_{(1)}}. $$ To do this, start with the right side of this equality and work toward the left side, not the other way around. I.e. start with $(Z_{(n)}-\bar{Z})/(Z_{(n)}-Z_{(1)}),$ not with $(X_{(n)}-\bar{X})/(X_{(n)}-X_{(1)}).$
Then show that the distribution of $Z_{(n)}, Z_{(1)},$ and $\bar Z$ does not depend on the value of $(\mu,\sigma),$ by using the fact that the distribution of the entire $n$-tuple $(Z_1,\ldots,Z_n).$