Suppose that $f: X \rightarrow Y$ is a morphism between algebraic varieties. If $Y$ is smooth, and the fibers of $f$ over closed points of $Y$ are proper and nonsingular, does it follow that $X$ is smooth?

Update: The answer to the question as posed, is NO. See a comment by Karl Schwede below for a counterexample.

Modified question: Let $f$ be a surjective morphism of algebraic varieties (reduced, irreducible, separated schemes over an algebraically closed field). Let $x \in X$ be a closed point and let $y = f(x)$. Just because the fiber $f^{-1}(y)$ is smooth does not mean $X$ is smooth at $x$. Is there any condition on $f$ or the fibers which will guarantee this? Flatness is one, is there any other condition?