To what extent can one imitate holomorphic functions in higher dimensions? Let $\Omega\subseteq \mathbb{C}$ be open. Let $f:\Omega \rightarrow \mathbb{C}$ be holomorphic. The last statement about $f$ can be restated as:
For every $z_0\in \Omega$, there exists $w\in \mathbb{C}$ such that:
$$\lim_{r\rightarrow 0^+}\sup_{\theta\in[0,2\pi]}\left|\frac{e^{-i\theta}[f(z_0+re^{i\theta})-f(z_0)]}{r}-w\right|=0$$
Now let $V\subseteq \mathbb{R}^3$ be open and $ f:V\rightarrow \mathbb{R}^3$ be a function. We say that $f$ is holomorphic iff for every $\vec{a}\in V$,  there exists $\vec{d}\in\mathbb{R}^3$ such that:
$$\lim_{r\rightarrow 0^+}\sup_{\text{R is a rotation}}\left\|\frac{R^{-1}(f(\vec{a}+rR(\vec e_1))-f(\vec{a}))}{r}-\vec{d}\right\|=0 \tag{1}$$
In $(1)$, $\vec e_1$ denotes the vector $(1,0,0)$.
Question:  Is my definition for holomorphic functions in 3D a "good" one ? For example, does it follow that $f$ is analytic ? Is there a standard name for functions that satisfy $(1)$ ?
Remark: There is a fact that (whose proof I don't know) that there is no topologically complete normed field that has $\mathbb{R}$ as a subfield other than $\mathbb{C}$. (I am not sure if I have the conditions of this fact right). This made me think that all the nice theorems of complex analysis that I have seen so far are special to $\mathbb{R}^2$. This makes me want to ask the question I asked above to see if complex analysis could be redone in $\mathbb{R}^n$ for $n\geq 3$. 
 A: Your proposal doesn't work very well. The problem with it is that -- in contrast to the 2D case -- knowing where $\vec e_1$ ends up does not uniquely pinpoint a rotation. Therefore for any given $\vec v$ there are manyl different rotations $R$ such that $\vec a+rR(\vec e_1)=\vec v$, and if you want to end up close to $f(\vec a)+r\vec d$ no matter which of their inverses you apply to $f(\vec v)-f(\vec a)$, that's a very strong condition on your $f$ can be.
Essentially your condition can only be met if $f$ is locally just a scaling with no rotation. If this is the case everywhere in an open subset of $\mathbb R^3$, $f$ must have the exact form $f(\vec x)=\vec c + k\vec x$ for some $k$ and $\vec x$. And your $\vec d$ is then $(k,0,0)$ everywhere.

The more common way to generalize differentiation to functions $f:\mathbb R^n\to \mathbb R^n$ is to allow the derivative to be an $n\times n$ matrix and say that the derivative of $f$ at $\vec a$ is $M$ iff
$$ \lim_{\vec s\to 0} \frac{|f(\vec a+\vec s)-(f(\vec a)+M\vec s)|}{|\vec s|} = 0 $$
(When this holds, $M$ is the Jacobian of $f$).
In two dimensions this gives rise to a more permissive concept than holomorphic complex functions, but we can recover the class of holomorphic functions by additionally requiring that $M$ must always be a nonnegative multiple of a rotation matrix. Then $f$ is a orientation-preserving conformal mapping, which happens to be the same as a holomorphic complex function if we identify $\mathbb R^2$ with $\mathbb C$.
In principle we can apply this same restriction to $n=3$, but Liouville's theorem says that the class of functions we get out of this is much less varied than the complex holomorphic functions.
