Question: Why a holomorphic function is infinitely differentiable just because of satisfying the Cauchy Riemann equations, but on the other side, a two variable real function that is twice differentiable is not infinitely differentiable?

I'm asking this for two reasons:

1) $\mathbb{R}^2$ is somehow an analog to the complex plane (they are isomorphic).

2) Holomorphic means that is differentiable in any direction of the complex plane, so why if a function is differentiable in any direction of the real plane then it is not infinitely differentiable?

Answer: 
The usual presentation of holomorphic functions (and complex analysis in general) is misleading; the analogue of a holomorphic function to a function on the real line is *one whose derivative is zero*!

Recall one nice property of holomorphic functions: if a point $p$ lies inside a region $D$, which has a boundary $C$, then you can find the value of a holomorphic function $f(p)$ by *only* a contour integral about the path $C$.

The analogue of this property to the real line is as follows: let there be a region (an interval) $D$, whose boundary is two points $C\_+$ and $C\_-$. Then the value of a function at a point $p \in D$ is characterized *entirely* by the value of the function at $C\_+$ and $C\_-$. The only way this is possible is if $f$ is *constant* on $D$ and has identical values at $C\_+$ and $C\_-$.

(Note: it's possible that $f$ has a delta function source in the region instead, but this is just the analogue of a meromorphic function.)

This generalization--functions whose derivatives are *zero*--is what is ultimately more useful when you start looking at vector calculus and higher-dimensional real spaces. Scalar fields that obey $\nabla \varphi = 0$ obey the exact same property of being fully characterized by their boundary values.

Similarly, the correct analogue on $\mathbb R^2$ is for a vector field $F$ to have vanishing divergence and curl: $\nabla \cdot F = \nabla \times F = 0$. Such vector fields also obey the same basic property of holomorphic functions.
Note that an analytic, complex function $w=f(z)$ implies a function $F:\mathbb R^2\rightarrow\mathbb R^2$ by $r+is=f(x+iy)$. The reverse is *not* true, however. It is *not* the case that just any differentiable $F:\mathbb R^2\rightarrow\mathbb R^2$ implies a complex, analytic function. On the contrary, the [Cauchy-Riemann equations](http://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations) must be satisfied. Thus, the real an imaginary parts are tightly tied together.