After reading the great answer in QiL's link, I decided to post it here specialized to the case I described.
Let $k$ be an algebraically closed field of characteristic zero and $B$ a finitely generated $k$-domain. We denote by $\Omega_{B/k}$ the module of differentials and by $d_B:B\to \Omega_{B/k}$ the universal derivation.
Claim: $\ker d_B=k$.
Proof: Let $L=Frac(B)$ and let $d_L:L\to \Omega_{L/k}$. We first prove the analogous result for $d_L$. Denote $E=\ker d_L$ and note that $E$ is a field (follows from Leibniz rule) and thus we have a tower of extensions $L/E/k$. The crutial point is that for separable extensions of fields (and here we use the assumption of zero characteristic) we have
$\mbox{tr.deg}(L/k)=\dim _L \Omega_{L/k}$
and
$\mbox{tr.deg}(L/E)=\dim _L \Omega_{L/E}$
On the other hand, we have a natural map $\Omega_{L/k} \to \Omega_{L/E}$ induced by the inclusion $k\to E$ (every $E$-derivation is in particular a $k$-derivation), but from the definition $E=\ker( {L\to \Omega_{L/k}})$ it is obvious that this map is an isomorphism (because every $k$-derivation is automatically an $E$-derivation) so we get
$\dim _L \Omega_{L/k}=\dim _L \Omega_{L/E}$
and together with the previous observation that
$\mbox{tr.deg}(L/k) = \mbox{tr.deg}(L/E)$
but this means that $\mbox{tr.deg}(E/k)=0$ and therefore that $E$ is algebraic over $k$ and since $k$ is algebraically closed it follows that $E=k$. What remains is to note that the inclusion $B\to L$ induces a natural map $\Omega_{B/k}\to \Omega_{L/k}$ so if $b\in B$ has zero differential then "chasing the diagram" it also has zero differential as an element of $L$ and therefore $b\in k$.
Remark: In the original proof there is also a reference to the question whether $\Omega_{B/k}\to \Omega_{L/k}$ is injective, but I don't see why is that important, at least in this situation, and I'll be happy to be corrected if I'am missing something.