Exercise III.10.2 in Hartshone's Algebraic Geometry is discussed in this question. The accepted answer proceeds by showing that the smooth locus is open on $X$, then transferring this to $Y$ using the fact that the morphism in question is closed.
As a commenter notes, it seems very strange that the full strength of properness hypothesis is not used. We are only using the fact that proper morphisms are closed, and not, for example, that they are universally closed or of finite type. Further, we are only supposing $X$ is locally Noetherian, a weaker hypothesis than being a variety (which the problem stipulates). However, I'm not able to find an error. Is this proof correct? Can we really get away with much weaker hypotheses, as the answer suggests?
(This question has been edited for clarity.)