Finding a function which is differentiable in specific points 
Find a function $\mathbb{C}\rightarrow \mathbb{C}$ s.t it will be differentiable just in the points $(1,1),(1,-1),(-1,1),(-1,-1)$

So we have to find a function $z=u(x,y)+iv(x,y)$ that satisfy $C-R$ equations with the given points:
$u_x=v_y$ and $u_y=v_x$
But where should I start? it seems that C-R equations should be something like $x^2$ so will "take" $\pm 1$
 A: Suppose $a,b,c,d$ are distinct points in  $\mathbb C.$ Then $f(z)=(z-a)^2(z-b)^2(z-c)^2(z-d)^2\chi_{\mathbb Q^2}$ is complex differentiable at each of $a,b,c,d$ and nowhere else.
Why? In $D(a,1)$ for example, $|f(z)|\le C|z-a|^2$ for some positive constant $C$ that would be too boring to present exactly. Anytime you have an estimate like that, the complex derivative will be $0$ at $a.$ Just go back to the definition of the complex derivative and it falls right out. The same of course applies to each of $b,c,d.$
At any point of $\mathbb C\setminus \{a,b,c,d\}$ you'll see that $f$ is not even continuous, much less differentiable, at that point. Of course we can take $a,b,c,d$ to be the given points in this problem, so we have an answer.
A: There is no such function, if a function is analytic at a point, it needs to be differentiable in a disk around that point, and then it is analytic in that entire disk.
If you want an example of a function which is differentiable at those four points, then here is a hint:
Hint $|z-z_0|^2$ is differentiable only at $z_0$ and has a zero derivative.
Hint 2: If $f$ is differentiable at $z_0$, with a zero derivative, and $g$ is  "not too bad" around $z_0$ then $fg$ has a zero derivative at $z_0$.
A: There are many options. The proposition by N.S. has the caveat that it introduces extra points where a complex derivative exist. For example, $|z-1|^2 |z+1|^2= |z^2-1|^2$ is in fact complex differentiable at $1,-1$ but also at $0$.
So you have to look for functions that avoids this problem. 
The easiest is often to look for a function for which the imaginary part vanishes identically, i.e. $v(x,y)\equiv 0$. Then you should find a real part which has critical points, i.e. points for which $u_x=0$ and $u_y=0$, precisely at the prescribed points.
Because of the symmetry of your points, one of the simplest possible choices in the present case could be  to take something like (your second C-R equation should btw be $u_y = - v_x$):
$$ u(x,y) = x^3-3x+ y^3-3y, \; v(x,y)=0.$$
One has $u_x=3(x^2-1)$ and $u_y=3(y^2-1)$ so that the C-R equations fail everywhere except at the given 4 points.
