Checking differentiability at the origin Let $f: \mathbb{R}^2 \to \mathbb{R} $ be given as 
$$
f(x,y) =
\begin{cases}
e^{\frac{-1}{x^2+y^2}}, & \text{if }\text{ $(x,y) \neq (0,0)$} \\
0, & \text{ }\text{ $(x,y)=(0,0)$}
\end{cases}
$$
I want to check $f$ is not differentiable at the origin. My attempt would be to consider the sequence $\mathbf{x_n} = (\frac{1}{n}, \frac{1}{n} ) \to (0,0) $. But, notice
$$ f( \mathbf{x_n}) = e^{\frac{-1}{2/n^2}}= e^{-\frac{n^2}{2}} \to e^0 = 1 \neq f(0,0) = 0$$
Hence, $f$ is not continuous at $(0,0)$. In particular, it cannot be differentiable at the origin. Is this a correct approach?
 A: The definition of differentiability can be stated in many ways.  The way that I prefer (and usually teach my students) is this:
Definition: We say that $f(x,y)$ is differentiable at $(x,y)=(a,b)$ if
$$
\lim_{(x,y)\to(a,b)}\frac{f(x,y)-L(x,y)}{\sqrt{(x-a)^2+(y-b)^2}}=0,
$$
where $L(x,y)$ is the linear approximation to $f$ near $(a,b)$:
$$
L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b).
$$
In your particular case, $(a,b)=(0,0)$. Using the definition of the partial derivative, you can show that $f_x(0,0)=f_y(0,0)=0$. This means that the linearization is
$$
L(x,y)=0+0(x-0)+0(y-0)=0.
$$
So, you need to check whether or not
$$
\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}\overset{?}{=}0.
$$
Since in the limit as $(x,y)\to(0,0)$, we can assume $(x,y)\neq(0,0)$, this is equivalent to checking whether or not
$$
\lim_{(x,y)\to(0,0)}\frac{e^{-1/(x^2+y^2)}}{\sqrt{x^2+y^2}}\overset{?}{=}0.
$$
Now, this limit is a perfect candidate for checking in polar coordinates:
$$
\lim_{(x,y)\to(0,0)}\frac{e^{-1/(x^2+y^2}}{\sqrt{x^2+y^2}}=\lim_{r\to0^+}\frac{e^{-1/r^2}}{r}\overset{?}{=}0.
$$
Can you see how to do this?
A: As Daniel Fischer commented, for your pair $(\frac1n,\frac1n)$ to approach $(0,0)$, you need $n\to\infty$, which makes $e^{-n^2/2}\to0$. Since that number is the same as the value at $(0,0)$, this does not show that the function is discontinuous at $(0,0)$.
Actually, the function is continuous at $(0,0)$. If you wanted to prove continuity, then you'd have to check more than just the sequence you described (which are only some of the points on the positive half of the line $y=x$), but account for (every sequence in) every path. If you know about polar coordinates, it can be done in a slick way with them. And if you don't but know how to work with, say, the $\varepsilon-\delta$ definition in this can be done more manually that way.
Since the function is continuous, you will have to use the definition of "differentiable" somehow. A multivariate function being differentiable at a point is a stronger condition than merely "the partial derivatives exist", or even "all directional derivatives exist", so if this doesn't sound familiar, you should look up the precise definition. 
A: Consider the function $\phi:\mathbb R\to\mathbb R$ defined by $\phi(0)=0$ and $\phi(t)=e^{-\frac1{\vert t\vert}}$ if $t\neq 0$. This function is differentiable at $t=0$ (with $\phi'(0)=0$) because $\frac{\phi(t)-\phi(0)}{t-0}=\frac{1}t\, e^{-\frac1{\vert t\vert}}\to 0$ as $t\to 0$ (recall that $ue^{-\vert u\vert}\to 0$ as $u\to\pm\infty$).
Now you have $f(x,y)=\phi(x^2+y^2)$. Since the function $\theta(x,y)= x^2+y^2$ is differentiable on $\mathbb R^2$ and $\theta(0,0)=0$, it follows that $f=\phi\circ\theta$ is differentiable at $(0,0)$ (with $Df(0,0)=0$).
