Much more is true: the general fibre is smooth. This is called *generic smoothness*; I will sketch a proof below.

Indeed, we may replace $X$ by the maximal open $U$ 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.

Then we have a dominant morphism $X \to Y$ of varieties. Let $\eta_X, \eta_Y$ be the generic points. Then $\kappa(\eta_Y) \to \kappa(\eta_X)$ is a field extension in characteristic zero, hence separable. Thus, $\Omega_{\kappa(\eta_X)/\kappa(\eta_Y)}$ is a $\kappa(\eta_X)$-vector space of dimension
\begin{align*}
r &= \operatorname{tr.deg} (\kappa(\eta_X)/\kappa(\eta_Y))\\
&= \operatorname{tr.deg} (\kappa(\eta_X)/k) - \operatorname{tr.deg} (\kappa(\eta_Y)/k)\\
&= \dim X - \dim Y.
\end{align*}
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 $\kappa(\eta_Y)$ is flat over $\kappa(\eta_X)$, 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.** What goes wrong in characteristic $p$ is that $\kappa(\eta_Y) \to \kappa(\eta_X)$ may be inseparable, in which case
$$\dim \Omega_{\kappa(\eta_X)/\kappa(\eta_Y)} > r.$$
The same proof as above shows that any¹ dominant morphism $X \to Y$ is generically smooth if and only if $\kappa(\eta_Y) \to \kappa(\eta_X)$ is separable.

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¹ For example, if $X$ and $Y$ are integral schemes and $f$ is of finite presentation. I don't know exactly what the most general situation is.