Originally Posted by
tonio You can evaluate the partial derivatives with the function as Mr. Fantastic showed you, or else realize (most books have it) that the CR-equations
are equivalent to $\displaystyle \frac{\partial f}{\partial\,\overline{z}}=0$ , where we have that $\displaystyle \frac{\partial}{\partial\,\overline{z}}=\frac{1}{2 }\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$ , so putting $\displaystyle z=x+iy$, we get:
$\displaystyle f(z)=f(x,y)=\frac{1}{1+x+iy}\Longrightarrow \frac{\partial f}{\partial x}=-\,\frac{1}{(1+x+iy)^2}\,,\,\,\frac{\partial f}{\partial y}=-\,\frac{i}{(1+x+iy)^2}$.
Check the above and conclude your function is analytic in $\displaystyle \mathbb{C}\setminus \{{\color{red}-1}\}$
Tonio