Cauchy-Riemann equation analogue but for the quaternions

Correct me if I'm wrong, but I believe the function you are describing is single variable: $$f:\mathbb{H}\to\mathbb{H}$$, which is represented by $$f:\mathbb{R}^4\to\mathbb{R}^4$$.

Starting from the method to derive the standard Cauchy Riemann equations, a function in a quaternion variable would be represented by $$f(q)=A_0+iA_1+jA_2+kA_3$$ where the $A_i$ are functions of four variables $x_i$ the differential can then be represented as $dq=dx_0+idx_1+jdx_2+kdx_3$, so $$\frac{df}{dq}=\sum_{i=0}^3\sum_{j=0}^3 \frac{\partial f}{\partial A_i}\frac{\partial A_i}{\partial x_j}\frac{\partial x_j}{\partial q}$$

$$=\left[\partial_{x_0}A_0+i\partial_{x_0}A_1+j\partial_{x_0}A_2+k\partial_{x_0}A_3\right]-\left[i\partial_{x_1}A_0-\partial_{x_1}A_1+k\partial_{x_1}A_2-j\partial_{x_1}A_3\right]-\left[j\partial_{x_2}A_0-k\partial_{x_2}A_1-\partial_{x_2}A_2+i\partial_{x_2}A_3\right]-\left[k\partial_{x_3}A_0+j\partial_{x_3}A_1-i\partial_{x_3}A_2-\partial_{x_3}A_3\right]$$ Since we assume that $f$ is differentiable, the derivative must have a single value as we approach a point from any line. We choose to approach the derivative from each of the axes, giving the equations $$\boxed{\frac{\partial A_0}{\partial x_0}\ =\ \frac{\partial A_1}{\partial x_1}\ =\ \frac{\partial A_2}{\partial x_2}\ =\ \frac{\partial A_3}{\partial x_3}\\ \frac{\partial A_0}{\partial x_1}\ =-\frac{\partial A_1}{\partial x_0}=-\frac{\partial A_2}{\partial x_3}=\ \frac{\partial A_3}{\partial x_2}\\ \frac{\partial A_0}{\partial x_2}\ =\ \frac{\partial A_1}{\partial x_3}\ =-\frac{\partial A_2}{\partial x_0}\ =-\frac{\partial A_3}{\partial x_1}\\ \frac{\partial A_0}{\partial x_3}\ =-\frac{\partial A_1}{\partial x_2}\ =\ \frac{\partial A_2}{\partial x_1}\ =-\frac{\partial A_3}{\partial x_0}}$$

As for your other questions, I don't know; I have not studied enough analysis to know what the theorem is.