Suppose $M$ is a differential manifold, and $C^\infty(M)$ denotes all the smooth functions on $M$. A linear operator $D:C^\infty(M) \rightarrow C^\infty(M)$ satisfies the Leibniz law: $D(f \cdot g)=D(f) \cdot g + f \cdot D(g)$. Must $D$ be the derivative alongside a given smooth vector field?
Bonus: What is the situation when we consider complex manifolds and holomorphic functions/vector fields?
For smooth manifolds, the answer is yes: indeed, it is trivial to see that $D$ is a rough vector field. To prove that it is a smooth vector field, let $D^i(x)\partial_i$ be a description of $D$ in local coordinates $(x_\alpha,U_\alpha)$. It is known that $D$ is smooth (on $U_\alpha$) iff $D^i$ is smooth for all $i$. Now let $\varphi$ be a bump function such that, outside the chart $U_\alpha$, $\varphi\equiv 0$ and $\varphi=1$ on a neighbourhood of $x_0\in U_\alpha$ fixed. Then, for every $j$, $D(x_{\alpha,j}\cdot \varphi)=D^j$. On the other hand, $D(x_{\alpha,j}\cdot \varphi)\in C^\infty(M)$ and so $D^j$ is smooth. Thus $D$ is a vector field locally smooth everywhere, which means it is smooth.
This proof breaks down for complex manifolds, since we do not have partitions of unity and or bump functions.