Ytukyg

Is the complex function $$f(z)=begin{cases}
e^{-z^{-4}} & zne0\
0 & z=0
end{cases}$$ 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 $(w_n)_{ngeq0}$ such that $w_ntoinfty$ and $exp w_n^4=1$ for all $ngeq0$. It follows that $1/w_nto0$ as $ntoinfty$ 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 $0$.

The function $f(z)$ thus defined has a Laurent series expansion
$$ f(z) = sum_{n=0}^infty frac{(-1)^n}{n!} z^{-4n} $$
around $z=0$ with an infinite principal part (i.e. with infinitely many non-zero coefficients in front of the negative powers of $z$).

Thus $z=0$ is an essential singularity of $f(z)$, and by the Casoratti-Weierstraß theorem, there is no way to define $f(0)$ so that $f(z)$ be continuous at $z=0$. In fact, we have
$$ |f(z)| = e^{-Re(z^{-4})} = expleft(frac{-x^4+6x^2y^2-y^4}{(x^2+y^2)^4}right) quad text{for} quad z = x +iy neq 0.$$
Restricting $z$ to the line $y = tx$ with $tinmathbb{R}$, we observe that
$$ lim_{xto0}|f(x+itx)| = lim_{xto 0},expleft(frac{-1+6t^2-t^4}{x^4(1+t^2)^4}right)tag{1}label{limit}$$
is not independent of $t in mathbb{R}$. Indeed, we can factor the biquadratic polynomial $-1+6t^2-t^4$ as
$$-big((t^2-3)^2 - 8big) = -big(t^2-(sqrt{2}-1)^2big)big(t^2-(sqrt{2}+1)^2big) = -(t+sqrt{2}+1)(t+sqrt{2}-1)(t-sqrt{2}+1)(t-sqrt{2}-1),$$
whence we deduce that the limit eqref{limit} equals either $1$ (for $t = pmsqrt{2}pm1$), or $infty$ (whenever $|t+sqrt{2}| < 1$ or $|t-sqrt{2}|<1$), or $0$ (otherwise). Therefore, we conclude that $lim_{zto0}|f(z)|$ does not exist. Hence, the limit $lim_{zto0}f(z)$ does not exist either, and thus, $f(z)$ cannot be analytic at $z=0$.

Edit: I would like to thank Abhishek Verma for pointing out an error in the original solution: the denominator appearing in the expression for $|f(z)|$ was incorrectly calculated as $(x^2+y^2)^2$, instead of $(x^2+y^2)^4$.

I would say no, because every derivative is $0$ at $0$, (due to expontential growth), see:
http://en.wikipedia.org/wiki/Non-analytic_smooth_function

Since you already know the function is continuous, we can conclude directly that it is analytic from Riemann's theorem (about removable singularity).