Show that $f$ is Gateaux differentiable

Question

Define a function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ as follows $$f(x,y)= \begin{cases} \frac{2y \exp(-x^{-2})}{y^2+ \exp(-2x^{-2})} & x \neq 0 \\ 0 & \text{otherwise} \end{cases} $$ Show that $f$ is Gateaux differentiable at $(0,0)$ but that $f$ is not continuous there. So I think I have to show that $f'((0,0),e_1)$ and $f'((0,0),e_2)$ exits and they are both linear. So far I have that $f'((0,0),e_1) =f'((0,0),e_2)=0$

Answer 1

Hint: Along a curve where $x^2 = -1/\ln y$ for $0< x,y < 1$, we have $f(x,y) = 1$. Hence $f$ is not continuous at $(0,0)$. The function is Gateaux differentiable if $\displaystyle \left.\frac{d}{d \epsilon}f(\epsilon h, \epsilon k)\right|_{\epsilon=0}$ exists for all $(h,k) \in \mathbb{R}^2$.

Note that in a neighborhood of $(0,0)$ with $y < 1$ and $x^2 = -1/ \ln y$ we find

$$f(x,y) = \frac{2y\exp(-1/x^2)}{y^2 + \exp(-2/x^2)}= \frac{2y^2}{2y^2 }= 1,$$

and

$$ \lim_{(x,y) \to (0,0); x^2 = -1/ \ln y}f(x,y) = 1 \neq 0 = f(0,0)$$