I have the following formulation of the Gateaux derivative for functions $f:\mathbb{R}^m\to\mathbb{R}^n$. Let $f(x) = \sum_{i=1}^n f_i(x)e_i$ where $e_i$ forms a basis for $\mathbb{R}^n$, and $f_i:\mathbb{R}^m\to\mathbb{R}$. Then $$ \begin{aligned} \text{d}f(u;\psi) & = \lim_{t\to 0}\frac{f(u+t\psi)-f(u)}{t} \\ & = \sum_{i=1}^n\left(\lim_{t\to0}\frac{f_i(u+t\psi)-f_i(u)}{t}\right)e_i \\ & = \sum_{i=1}^n(\nabla f_i\cdot\psi)e_i \end{aligned} $$ From this it is easy to see that $\text{d}f(u;\psi)$ is linear in $\psi$. Furthermore, it would seem that the existence of the Gateaux derivative is equivalent to the existence of the partial derivatives of $f_i$, which is also what P. Drabek and J. Milota state in their book "Methods of Nonlinear Analysis" (p. 118). However, it appears that not only linearity is "false", but equivalence of existence is false as well. Consider $m=2$ and $n=1$, and $$ f(x,\ y) = \frac{x^3}{x^2+y^2} $$ with $f(0,\ 0) = 0$. The partial derivatives do not exist at $0,\ 0$, as the limit is path dependent, but the Gateaux derivative does exist.
This, I can mentally justify. The partial derivatives do exist along a certain path (say, $(x, y)\to(0,\ 0)$ as $(at, bt)\to(0,\ 0)$), so that's rectified. However, as the Wikipedia article for Gateaux derivatives states, $$ \text{d}f(0,0;a,b) = \frac{a^3}{a^2+b^2} $$ (and $0$ at $a=b=0$). This function however is not linear. So basically my question is: what is happening? Part of my confusion must lie in the fact that it's 3am where I live, but I've thought about this for a while and I just can't seem to rectify it in my head. Why am I getting this conflict?