As pointed out by potentially dense below, this argument does not work in full generality. However, it does work when either $X$ is smooth, or the general fibre has dimension $0$.

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.

> **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.* Let $\eta_X, \eta_Y$ be the generic points, and write $K(X) = \kappa(\eta_X)$, $K(Y) = \kappa(\eta_Y)$ for the function fields of $X$ and $Y$. Then $K(Y) \subseteq K(X)$ is a field extension in characteristic zero, hence separable. 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.** 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. However, I think that under the assumption that $K(Y) \subseteq K(X)$ is separable, it should still be ok.

**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 $k[T]/(T^p - \alpha) = k[T]/((T-\alpha^{1/p})^p)$.