It does however not follow that such functions need be analytic or even $C^{2}$. Therefore the hyperbolic CR operator perhaps doesn not impose as much structure on functions it annihilates as does its complex counterpart, but it gives some structure none the less. **please forgive me for consistantly failing to write this down correctly: what I wanted to say was that the CR operator is equal to $\frac{1}{2}(\frac{\partial}{\partial x}+j\frac{\partial}{\partial y})$.

Also if one defines differentiability for functions in $\mathbb{R}[j]$ as the existance of such $f'(z)$ for every direction $$\lim_{h\to 0}\frac{f(z+h)-f(z)-f'(z)h}{\lVert h\rVert}=0$$ Here $\lVert h\rVert$ is the standard euclidean norm on $\mathbb{R}^2$, using which prevents the problem of zero devisors with the definition of derivatives. Notice that the condition that $f'(z)$ exists is the same as requiering CR operator to annihilate the function at $z$, simmilarly as in complex analisys.

...to the form $\frac{1}{2}(\frac{\partial}{\partial x}-j\frac{partial}{\partial y})$. Your account is incorrect because of the following fact: consider the operator $$\frac{1+j}{2}\frac{\partial}{\partial (x+y)}+\frac{1-j}{2}\frac{\partial}{\partial (x-y)}$$ The function annihilated by it are the hyperbolic equivalent of antiholomorphic.

The structure $\mathbb{R}[x]/(x-1)$ corresponds to $z=x+jy$ for $x,y$ real. In this notation zero divisors are percisely those numbers with $x^2-y^2=(x+y)(x-y)=0$. Than you can construct a basis alternative to $\{1,j\}$ from zero divisors $\{(1+j)/2,(1-j)/2\}$ so that $$z=x+jy=\frac{1+j}{2}(x+y)+\frac{x-y}{2}(x-y)$$ Now in this form, the CR operator indeed gets the form $\frac{1+j}{2}\frac{\partial}{\partial (x-y)}+\frac{1+j}{2}\frac{\partial}{\partial (x-y)}$ which can easily be proven equivalent to the form $\frac{1}{2}\Big (\frac{\partial}{\partial x}-j{\partial}{\partial y}$