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Working locally on $X$ and $Y$, we may assume they are affine and so the map $f : X \to Y$ corresponds to a ring map $S \to R$ (an inclusion) with $S$ smooth over the base field and $R$ normal. Then the generic fiber is simply $(S \setminus 0)^{-1} R$. That's certainly normal since a multiplicative set times a normal ring is still normal. It easily follows that an open set of the other fibers are also normal.