Problem
In which region of the complex plane is the following function analytic?
$$f(z) = \dfrac{z^{2}}{\mathrm{e}^{x} \cos y+i \mathrm{e}^{x} \sin y}$$
If the function has a derivative over its domain, determine $f'(z)$.
Progress
First I expressed $f (z)$ as $u (x, y) + v (x, y).$
$$f(z) = \left[ \dfrac{x.\cos y + y.\sin y}{\mathrm{e}^x} + \left( \dfrac{y\cos y + x\sin y}{\mathrm{e}^x} \right) i \right],$$
$$f(z) = u(x,y) + v(x,y).$$
Then I used the Cauchy-Riemann equations
$\dfrac{\partial u}{\partial x} = -\left(\cos\left(y\right)x+y\sin\left(y\right)-\cos\left(y\right)\right)\mathrm{e}^{-x}$
$\dfrac{\partial u}{\partial y} = \mathrm{e}^{-x}\left(-x\sin\left(y\right)+\sin\left(y\right)+y\cos\left(y\right)\right)$
$\dfrac{\partial v}{\partial y} = \mathrm{e}^{-x}\left(-y\sin\left(y\right)+x\cos\left(y\right)+\cos\left(y\right)\right)$
$\dfrac{\partial v}{\partial x} = -\left(\sin\left(y\right)x-\sin\left(y\right)+y\cos\left(y\right)\right)\mathrm{e}^{-x}$
then
$\begin{aligned} &\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \\ &\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y} \end{aligned}$
But I don't really understand the rationale for using the Cauchy-Rieman equations in this problem, and I can't determine what the region is. I am trying to see it from the most essential aspect (using epsilon and delta, and compact sets) but I don't understand.
Your functions $u$ and $v$ are (real) differentiable and satisfy the Cauchy-Riemann equations on the whole plane. So the function $f$ is analytic.
To find $f'(z)$, even if you don't notice that $f(z)=z^2e^{-z}$, you can just use: $f'(z) = u_x-iu_y$.