I made up this rule and I think it should work:
$$\lim_{h\rightarrow 0}\frac{f(z+h(\cos\theta+i\sin\theta))-f(z)}{h(\cos\theta+i\sin\theta)}$$
If the limit exists and is independent of $\theta$, then the function is holomorphic or differentiable at $z$.
Clearly, fixing a particular value of $\theta$ means approaching $z$ from one particular direction. So, if this limit is independent of $\theta$, then the function has the same derivative from whatever direction we approach it. Is this rule correct?
Consider the function $f(x) = 0$ for $x\in \mathbb{R}$ and $f(re^{i\theta'}) = \frac{1}{\theta'}r^2$, where $r > 0, \theta' \in (0,2\pi)$. Check: Your limit for $z=0$ is 0 regardless of the choice of $\theta$, but $f$ is not even continuous at $0$, as $f$ takes arbitrarily large values in any neighbourhood of $0$.