Is the complex function $f(z)=e^{-z^{-4}}$ analytic at $z=0$? Is the complex function $$f(z)=\begin{cases}
e^{-z^{-4}} &amp; z\ne0\\
0 &amp; z=0
\end{cases}$$ analytic at $z=0$ ?

I am able to show that the Cauchy-Riemann equations are satisfied.
The other answers are using thermonuclear weapons to conclude...

The function is not even continuous at zero. For example, there exists in $\mathbb C$ a sequence $(w_n)_{n\geq0}$ such that $w_n\to\infty$ and $\exp w_n^4=1$ for all $n\geq0$. It follows that $1/w_n\to0$ as $n\to\infty$ and $f(1/w_n)=\exp(-(1/w_n)^{-4})=1$ for all $n$, and this together with $f(0)=0$ imply that $f$ is not continuous at $0$.
Since you already know the function is continuous, we can conclude directly that it is analytic from Riemann's theorem (about removable singularity).