I have the following problem:
Let $f:\mathbb{R}^2 \to \mathbb{R}$ be defined by:
\begin{equation} f(x,y)= \frac{x^3y}{x^4+y^2}, \quad x \neq 0, y\neq 0 \end{equation} and: \begin{equation} f(x,y)=0, \quad x=0,y=0 \end{equation} Show that $f$ is Gateaux differentiable at $0$ and the Gateaux derivative at that point is $0$. Moreover, show that $f$ is not Frechet differentiable at $(0,0)$.
These concepts of differentiability are new to me, therefore all I need is a detailed way of proving what this problem asks for. Thank you all for your help!
Let $(u,v)\neq(0,0)$.
$D_{(0,0)}f(u,v)=\displaystyle\lim_{\epsilon\to0}\dfrac{f\big((0,0)+\epsilon((u,v))\big)-f(0,0)}{\epsilon}=\lim_{\epsilon\to0}\dfrac{f(\epsilon u,\epsilon v)}{\epsilon}=\lim_{\epsilon\to0}\dfrac{\epsilon^4 u^3v}{\epsilon(\epsilon^4 u^4+\epsilon^2 v^2)}=\lim_{\epsilon\to0}\dfrac{\epsilon^4}{\epsilon\cdot\epsilon^2}\left[\dfrac{u^3v}{\epsilon^2 u^4+v^2}\right]=0$
(Note that if $(u,v)=(0,0)$ we obtain the same)
Then, $D_{(0,0)}f=\begin{pmatrix}0\\ 0\end{pmatrix}$.
But $f$ cannot be Fréchet differentiable because has a removable discontinuity at $(0, 0)$ (one can see by approaching the origin along the curve (t,t)).