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). Now the result follows from Tag 0578. $\square$
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, 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 that $X$ is smooth. Then the general fibre of $f$ is smooth.
Proof. Since $X$ is smooth, 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 $K(Y)$.
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. If we replace smoothness of $X$ by the condition that $f$ is generically finite, then $X_{\eta_Y}$ is just $\{\eta_X\}$, which is smooth over $\{\eta_Y\}$ since it is a separable field extension. We can then finish the argument in the same way.
But for $0$-dimensional fibres, being smooth is not stronger than being geometrically reduced, so this only gives an alternative proof for the lemma above in this special case.
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 problem is that smooth is not equivalent to regular (but to geometrically regular) over imperfect fields, so if $X$ is smooth, we cannot conclude that $X_{\eta_Y}$ is smooth over $K(Y)$. See also the example above.
However, I think that under the assumption that $K(Y) \subseteq K(X)$ is separable, it should still be ok.
