where is this function with exponential differentiable? I need to find out where the function
$$f(x,y)=\left\{
\begin{array}{cc}
    e^{\tfrac{1}{x^2+y^2-1}}, & x^2+y^2<1  \\
    0,  & x^2+y^2\geq1
\end{array}
\right.$$
is differentiable. Here is my progress
For $x^2+y^2<1$, I can use the chain rule
$$\frac{\partial f}{\partial x}=\frac{-2xe^{\tfrac{1}{x^2+y^2-1}}}{(x^2+y^2-1)^2}, \ \ \frac{\partial f}{\partial y}=\frac{-2ye^{\tfrac{1}{x^2+y^2-1}}}{(x^2+y^2-1)^2}.$$
This partial derivatives are continiuos, so $f$ is differentiable.
For $x^2+y^2>1$, $f$ is constant, then its partial derivatives are zero and $f$ is differenciable in this case too.
Finally, for $x^2+y^2=1$, the fraction $\dfrac{1}{x^2+y^2-1}$ goes to $0$ and $f(x,y)=1$ and we have
$$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0.$$
Therefore, $f$ is differentiable in the whole $\mathbb{R}^2$.
Is it correct?

Let's consider $$f(t)=\left\{
\begin{array}{cc}
    e^{\dfrac{1}{t-1}}, & t<1  \\
    0,  & t\geq1
\end{array}
\right.$$
Which is differentiable and its composition with differentiable $g(x,y)=x^2+y^2$
For partial derivative let's consider any $(x,y)$ for which $x^2+y^2=1$. Then
$$\lim_{\Delta x \to 0-}\frac{e^{\frac{1}{(x+ \Delta x)^2+y^2-1}}}{\Delta x}=\lim_{\Delta x \to 0-}\frac{1}{\Delta x}e^{\frac{1}{(x +\Delta x)^2- x^2}}=\lim_{\Delta x \to 0-}\frac{1}{\Delta x} e^{\frac{1}{2 x \Delta x+\Delta x^2}}=0$$
