In differential geometry, we know that given a smooth map between smooth manifolds $\phi:X\to Y$, such that $X$ is connected and $d\phi|_x\equiv0$ for all $x\in X$, then $\phi$ is constant (this follows directly from the fact that a real function on a connected domain whose derivative is identically zero is constant).
Now, we can ask whether the same is true about algebraic varieties. i.e where $X,Y$ algebraic varieties over some algebraically closed field $k$ and $\phi$ a morphism of algebraic varieties (with the differential interpreted suitably). Well, the general answer is NO with the canonical counterexample being the Frobenius morphism $x\mapsto x^p$ for $\mathbb{A}^1\to \mathbb{A}^1$ when $\mbox{char}(k)=p$ (it's not constant, but it has zero differential).
It seems though that in zero characteristic this problem can't happen, but I would like to see a proof of that (elementary as possible, even computational) or a counterexample.
My best try: Using standard arguments this can be reduced to the case where $X$ is irreducible affine and $Y=\mathbb{A}^1$. Embedding $X$ in $\mathbb{A}^n$ and translating to algebra we get the following formulation:
Let $R=k[x_1,...,x_n]$ , $f\in R$ and $P=(g_1,...,g_r)$ a prime ideal in $R$. Suppose that for all $a\in k^n$ such that $g_1(a)=...=g_r(a)=0$ we have $df|_a\in \mbox{Span}(dg_1|_a,...,dg_r|_a)$, then there is $c\in k$ such that $f-c\in P$.
It is known (though still not trivial) that if we had $df|_a\equiv0$ for all $a$ in the mutual zero set of $g_1,...,g_r$, which is equivalent to saying that all the partial derivatives of $f$ are in $P$, then we would have the desired conclusion. Perhaps there is a way to reduce the problem to this case.
Remark: This is a fairly natural question in my opinion and I was surprised that I couldn't find any reference to it in my algebraic geometry books or just by googling it.