We may replace $X$ by the maximal open on which $f$ is defined (I am assuming that this is what you mean by the fibre of a rational map). If $f$ is not dominant, then the general fibre is empty, and the result vacuously holds. So we may assume $f$ is dominant.

Let $\eta_X$ (resp. $\eta_Y$) be the generic point, and let $K(X) = \kappa(\eta_X)$ (resp. $K(Y) = \kappa(\eta_Y)$) be the function field of $X$ (resp. $Y$).

> **Lemma.** Let $k$ be a field of characteristic $0$, and $f \colon X \to Y$ a dominant morphism of integral $k$-varieties. Then the general fibre of $f$ is reduced.

*Proof.* The generic fibre is integral since it is the localisation $X_{\eta_Y} = X \times_Y K(Y)$ (for rings: if $A \subseteq B$ are domains, then $B \otimes_A \operatorname{Frac} A \subseteq \operatorname{Frac} B$ is a domain).

Since $K(Y)$ is perfect, the reduced $K(Y)$-variety $X_{\eta_Y}$ is geometrically reduced (see [Tag 020I][1]). Now the result follows from [Tag 0578][2]. $\square$

However this is false in characteristic $p$, even over an algebraically closed field:

**Example.** Let $k$ be algebraically closed of characteristic $p > 0$, and $X \to Y$ given by the algebra morphism
\begin{align*}
\phi \colon k[T] &\to k[T]\\
T &\mapsto T^p.
\end{align*}
Then the fibres of $f$ have dimension $0$, and the generic fibre is reduced (but not geometrically reduced, let alone smooth). However, *every* closed fibre is nonreduced: for $\alpha \in k$, the fibre over $(T - \alpha)$ has coordinate ring $k[T]/(T^p - \alpha) = k[T]/((T-\alpha^{1/p})^p)$.

------------

**Remark.** If $X$ is smooth or $f$ is generically finite (i.e. the general fibre is $0$-dimensional), then much more is true. (This was my original answer, but it contained a serious mistake, as pointed out by potentially dense in the comments.)

> **Theorem.** (Generic smoothness) Let $f \colon X \to Y$ be a dominant morphism of varieties over a field $k$ of characteristic $0$. Assume either that $X$ is smooth, or that the general fibre has dimension $0$. Then the general fibre of $f$ is smooth.

*Proof.* The field extension $K(Y) \subseteq K(X)$ is separable, since we are in characteristic $0$. Thus, $\Omega_{K(X)/K(Y)}$ is a $K(X)$-vector space of dimension
\begin{align*}
r &= \operatorname{tr.deg} (K(X)/K(Y))\\
&= \operatorname{tr.deg} (K(X)/k) - \operatorname{tr.deg} (K(Y)/k)\\
&= \dim X - \dim Y.
\end{align*}
We have two cases:

 - If $X$ is smooth, then all local rings $\mathcal O_{X, \xi}$ are regular. If $\xi \in X_{\eta_Y}$ is a point in the generic fibre, then the local ring $\mathcal O_{X_{\eta_Y}, \xi}$ equals $\mathcal O_{X, \xi}$. Indeed, if $A \subseteq B$ is a map of domains and $\mathfrak p \subseteq B$ is a prime ideal with $\mathfrak p \cap A = (0)$, then $B_{\mathfrak p} = (B \otimes_A \operatorname{Frac} A)_{\mathfrak p}$.
Thus, we conclude that $X_{\eta_Y}$ is regular, hence smooth over $\kappa(\eta_Y)$.

 - If the general fibre has dimension $0$, then so does the generic fibre. Hence $X_{\eta_Y}$ is just $\{\eta_X\}$, so the generic fibre is smooth.

Thus, in both cases the generic fibre is smooth. This means that $\Omega_{X_{\eta_Y}/K(Y)}$ is locally free of rank $r$. Hence by the localisation properties of $\Omega$ there is an open set $V \subseteq Y$ such that $\Omega_{f^{-1} V/V}$ is locally free of rank $r$. Similarly, since $X_{\eta_Y}$ is flat over $K(Y)$, there exists an open $V' \subseteq Y$ over which $f$ is flat. Taking $W = V \cap V'$, we find that
$$f \colon f^{-1}(W) \to W$$
is smooth of relative dimension $r$. $\square$

**Remark.** We can weaken smoothness of $X$ to the assumption
$$\dim X^{\operatorname{sing}} < \dim Y.$$
Indeed, if this is the case we may remove from $Y$ the image of $X^{\operatorname{sing}}$ to reduce to the case where $X$ is smooth.

**Remark.** What goes wrong in characteristic $p$ is that $K(Y) \subseteq K(X)$ may be inseparable, in which case
$$\dim \Omega_{K(X)/K(Y)} > r.$$
Another potential problem is that smooth is not equivalent to regular (but to *geometrically regular*) over imperfect fields, so in the case that $X$ is smooth, we cannot conclude that $X_{\eta_Y}$ is smooth over $K(Y)$. However, I think that under the assumption that $K(Y) \subseteq K(X)$ is separable, it should still be ok.


  [1]: http://stacks.math.columbia.edu/tag/020I
  [2]: http://stacks.math.columbia.edu/tag/0578