Continuity and partial differentiability for $f(x, y) := e^x - 1$ 
Given
$$M =\left\lbrace(x, y) \in \Bbb R^2 \setminus 0 \ \mid\ x = y\right\rbrace$$
and
$$f(x, y) := e^x - 1$$
when $(x, y) \in M$,
otherwise
$$f(x, y) := 0,$$
I have to examine whether the function is continious and partial differentiable at $(x, y) = 0$ or not.

This exercise looks a little bit weird to me since it was part of an exam and since it looks a little bit too easy to me. I might've done something wrong.
Approach
First, we note that
$$(x, y) = 0\quad \implies \quad(x, y) = (0, 0) \quad\implies\quad (x, y) \in M.$$
Continuity
We can simply plug the values into the function:
$$\lim_{(x,y)\to (0,0)} e^x - 1 = \lim_{x\to 0} e^x - 1 = e^0 - 1 = 1 - 1 = 0.$$
Therefore, the function is continious at $(0, 0).$
Partial differentiability
$$\lim_{{(0,y)}\to (0,0)}  e^x - 1 = e^0 - 1 = 1 - 1 = 0,$$
and similar
$$\lim_{(x, 0)\to 0} e^x - 1 = 0.$$
Since the critical values do exist in both cases, the function is partial differentiable at $(x, y) = (0, 0).$
 A: There is a lot of nonsense going on here.
In the first place, the function $f$ is defined on all of ${\mathbb R}^2$. It is $=0$ everywhere, except on the set $M:=\{(x,y)\>|\>x=y\ne0\}$. For $(x,y)\in M$ it is stipulated that $f(x,y):=e^x-1$.
The function $f$ is continuous at all points $(x,y)\notin M\cup\{{\bf 0}\}$, because it is $\equiv0$ in a full neighborhood of such points. If $(x_0,y_0)\in M$ then $f(x_0,y_0)\ne0$, but $f(x,y)=0$ at points $(x,y)$ arbitrarily close to $(x_0,y_0)$. It follows that $f$ is discontinuous at all points of $M$.
Remains the point ${\bf 0}$. I claim that $f$ is continuous there. Proof: The exponential function is continuous. Hence, given an $\epsilon>0$, there is a $\delta>0$ such that $|e^x-1|<\epsilon$ whenever $|x|\leq\delta$. Assume now that $(x,y)\in B_\delta({\bf 0})$. If $(x,y)\notin M$ then $f(x,y)=0$. If $(x,y)\in M$ then from $|x|\leq\delta$ it follows that $|f(x,y)|=|e^x-1|<\epsilon$.
Since $f$ is $\equiv0$ on both the $x$- and the $y$-axis it is obvious that
$${\partial f\over\partial x}(0,0)={\partial f\over\partial y}(0,0)=0\ .$$
