I solved this exercise, but I am insecure of myself so I would like some comment from you about the correctness, or not, of my reasoning.
Consider the set $A$ and the function $f(x, y)$ such that
$$A = \{(x, y)\in\mathbb{R}^2; x > 0, y < e^{-1/x}\}$$
$$f(x, y) = \begin{cases} 2x & (x, y) \in A \\\\ y & (x, y) \not\in A \end{cases}$$
The question is: for which points of $\partial A$ is the function continuous?
Proceedings
First of all I had to find the boundary of $A$. I drew it on a paper, with shot consierations about the function $e^{-1/x}$. Since I have $x > 0$, at the end the boundary is given by
$$\partial A = (x, y) \in \mathbb{R}^2; (x > 0, y = e^{-1/x}) \cup (x > 0, y = 0)$$
(Notice: the second set should represent the equation of the positive $x$-axis, but I don't know if that is the correct way to write it.)
Now the only way for the function to be continuous is that $2x = y$ for certain $(x, y)$. This means
$$2x = y \rightarrow x = \frac{y}{2}$$
The only points belonging to the boundary satisfying this conditions are on the positive $x$-axis at $(x, y) = (x, y/2)$ (which are infinite).
So, how much wrong am I and where? Thank you!
$$\color{red}{\delta A}=\{x= 0, y\le 0\}\cup \{x\ge 0,y=e^{-\frac 1x}\}$$ Using desmos
The continuity of f needs checked around $\delta A$
$$\text{Continuity around (0,y) }, y\lt 0: \lim_{(x,y)\to (0,y)}{f(x,y)}=\lim_{x\to 0^-}{y}=\lim_{x\to 0^+}{2x}=\underbrace{y}_{f(0,y)}\Rightarrow (x,y)=(0,0):\text{ f discontinuous}$$
$$\text{Continuity at }(0,0):\lim_{(x,y)\to (0,0)}{f(x,y)}=\lim_{y\to 0}y=\lim_{x\to 0}{2x}=\lim_{x\to 0}{e^{-\frac 1x}}=0, \text{ f continuous at (0,0)}$$
$$\text{Cmotinuity around }y=e^{-\frac 1x}, x\gt 0:y=2x=e^{-\frac 1x}\Rightarrow (x,y) \notin \Bbb R^2, \text { f discontinuous}$$
$$\text{ f continuous on }\delta A \text{ at } (0,0)(\therefore)$$