Is the empty function differentiable?

Yes, the empty set is a smooth manifold (it is covered by the empty collection of coordinate charts!). It has every dimension. (That is, for any $n$, it is true that $\emptyset$ is a manifold of dimension $n$. Note that there is not really a unified definition of "manifold" but rather a separate definition of "manifold of dimension $n$" for each $n$, so there is nothing wrong with a single object satisfying the definition for multiple different values of $n$.)

For any smooth manifold $X$, the empty function $\emptyset\to X$ is smooth. After all, this just means that it gives a smooth function in every pair coordinate charts of a pair of atlases on the domain and codomain, which is vacuously true since the empty set is an atlas for $\emptyset$. (Or, in the context of just open subset of Euclidean space, if we consider $\emptyset$ as an open subset of $\mathbb{R}^m$, then the empty function $\emptyset\to\mathbb{R}^n$ is smooth because it is vacuously infinitely differentiable at every point in the domain. Its derivative is then the empty function $\emptyset\to \mathbb{R}^{n\times m}$.)

Yes, the empty function is differentiable because at
every point in the empty set, it is differentiable.