If $f:X \rightarrow S$ is locally of finite type, there is unique largest open subset $U$ in $X$ such that $f|U$ is etale. 


Suppose $f$ is finite and $U$ is nonempty. Is it true that $f|U$ is finite etale?


Thanks in advance.



Restricting finite morphism to finite etale morphim
No, because $U\to S$ finite implies that $U\to X$ is finite (at least when $X\to S$ is separated), so $U$ would be closed in $X$. 


If you want an example, take a non-trivial morphism from a projective smooth curve $X$ to the projective line over $\mathbb C$. 


You might ask whether $U\to f(U)$ (if the latter is open in $S$) is finite, but this is not true either. 
Consider for example $S=\mathrm{Spec}\mathbb Z$ and $X=\mathrm{Spec}\mathbb Z[t]$ with $t^3+t^2+2t+2=0$. 
In the fiber above $2$, there is one étale point and one ramified point.


