Question about $f(z)=\exp (-\frac{1}{z^4})$ Let $f(z)=\exp (-\frac{1}{z^4})$ for $z\neq 0$ and $f(0)=0$.
I know this is a famous example and got asked a lot
However my question is not about the origin, but what is the best way to actually show that this functions satisfies Cauchy Riemann everywhere? I only read  that it fulfills it everywhere but nowhere could I find an actual proof.
Because writing
$ z = x+iy$ and then expanding the denominator in the exponent and then dividing into real and imaginary part ist really exhausting. Is this the only way or is there some shortcut here?
 A: We rarely work with definitions alone; we use all the theorems available at our disposal. If I asked you to prove that $f(x)=e^{-1/x^4}$ is differentiable on $\Bbb{R}\setminus\{0\}$, you had better tell me that the reason (in full) this is true is that:

*

*$x\mapsto x$ is differentiable (obvious from $\epsilon$-$\delta$)

*$x\mapsto x^4$ is thus differentiable, being a four-fold product of differentiable functions

*$x\mapsto \frac{1}{x^4}$ is differentiable away from the origin (the quotient of continuous functions is differentiable away from the zeros of the denominator)

*$x\mapsto -\frac{1}{x^4}$ is differentiable away from the origin (product of the constant function $-1$ with a differentiable function)

*$\exp:\Bbb{R}\to\Bbb{R}$ is differentiable (a well-known fact)

*$x\mapsto e^{-1/x^4}$ is differentiable away from the origin (composition of differentiable functions is differentiable).

In the complex case $f:\Bbb{C}\setminus\{0\}\to\Bbb{C}$, $f(z)=e^{-1/z^4}$, the reasoning above works word for work once you replace $\Bbb{R}$ with $\Bbb{C}$. The sum, product, quotient, chain rules (compositions) are all stated and proved exactly as in the real case. So, no Cauchy-Riemann necessary here (and this is often an extremely inefficient way of doing things when you’re given nice simple explicit formulas $f(z)=\dots$, for exactly the reasons you mentioned). Once you know that the function is complex-differentiable, you can then state as a simple corollary that it must therefore satisfy the Cauchy-Riemann equations.
A: As I think there is a bit of confusion about complex differentiability vs C-R and there are lots of subtle points, let's note that if we have $f: U \to \mathbb C$ a complex differentiable function in an open (domain) $U \subset \mathbb C$, then by the usual theorems, $f$ is infinitely differentiable, analytic and satisfies the C-R equations which are succintly written as $$\frac{\partial f}{\partial y}=i\frac{\partial f}{\partial x}$$
However, the converse is not true in the sense that if $f$ is defined in $U$, has partial derivatives $\frac{\partial f}{\partial y},\frac{\partial f}{\partial x}$ everywhere in $U$ and said partial derivatives satisfy the C-R above, then it does not follow that $f$ is complex differentiable in $U$ and $f(z)=\exp (-\frac{1}{z^4}), z \ne 0$, while $f(0)=0$ is such a counterexample as clearly $f$ is complex differentiable on $\mathbb C-0$ being a composition of complex differentiable functions there, so by part I above it does satisfy C-R there.
Now it is a straightforward exercise to show that  $\frac{\partial f}{\partial y}(0)=\frac{\partial f}{\partial x}(0)=0$ so indeed $f$ has partial derivatives everywhere in the plane, satisfies C-R everywhere too, but it is not complex differentiable everywhere and there are counterexamples where the points where such a function (partial derivatives everywhere, C-R everywhere) is not complex differentiable have non zero Lebesgue measure.
(Note that $f$ above is separately continuous and real differentiable at $0$ in $x$ and $y$ as $f(x+i0)=e^{-1/x^4}$ which is the typical $C^{\infty}$ but not real analytic function, while $f(0+iy)=e^{-1/y^4}$ also)
In usual CA texts, one assumes stronger hypothesis like $f$ real differentiable with continuous differential on an $U$ plus C-R or at least $f$ continuous and the partial derivatives are continuous on $U$ too and deduce that $f$ is complex differentiable on $U$, but there is a theorem of Looman-Menchoff (strengthened by Montel) which is quite difficult and subtle (took a few decades to realize that the original proof of Looman had a gap and fix it) which says that $C-R$ plus continuity on $U$ (or just local boundness - Montel) are enough for complex differentiability as the $f$ here clearly fails continuity or boundness at the origin.
There is a very informative paper by Gray and Morris discussing this and much more:
"When is a Function that Satisfies the Cauchy-Riemann Equations Analytic?"
