$G = \frac{|x|^{\alpha} y^{1/3}}{|x|+|y|}$, when $(x,y)\neq (0,0)$ and $0$ when $(x,y) = (0,0)$.
For which $\alpha$ does $G'(0;\mathbf h)$ (the directional derivative of $G$ along $\mathbf h$) exist ($\forall \mathbf h\in \mathbb R^2$)?
According to the definition of directional derivative it's equal to the following limit ($h=(a,b)$):
$$\lim_{t\to 0}\frac{G(t\mathbf h)}{t} =\lim_{t\to 0} \frac{|ta|^\alpha b^{1/3}}{t(|a|+|b|)}$$
So, the exercise could be solved by showing for values of $\alpha$ for which the limit exists. I think it doesn't exist, so the directional derivative never exists, because the limit of $|t|^\alpha/t$ doesn't exist for any $\alpha$. Am I correct?