What is $d\bar z$ as a form? If $\omega \in F^r$ is complex r-form then we can define complex conjugate element $\bar\omega \in F^r$ by setting 
$\bar\omega(v_1,.....v_r)= \overline{\omega(v_1,......,v_r)}$ where $v_1,\ldots,v_r$ are coming from tangent space.
So if $\omega$ is $(p,q)$ form then $\bar\omega$ is $(q,p)$ form.
Now $dz$ is $(1,0)$ form so $d\bar z= \overline {dz}$ is $(0,1)$ form.
Hence $d\bar z(\frac {\partial}{\partial{\bar z}})=\overline {dz(\frac {\partial}{\partial{\bar z}})}=\bar 0=0$.
But it should be $1$ as $d\bar z$ is element of dual basis w.r.t. basis $(\frac {\partial}{\partial{z}},\frac {\partial}{\partial{\bar z}})$.
Then what mistake I am making?
 A: In Fritzsche & Grauert's definition of the complex conjugate of a form, they are assuming $v_1,\dots,v_r$ are elements of the tangent space, meaning they are real vectors. If you allow them to be complexified tangent vectors, then you have to write
$$\bar\omega(v_1,.....v_r)= \overline{\omega(\bar v_1,......,\bar v_r)}.$$
If you insert this into your computation, you'll see that it fixes the problem.
Added: Fritzsche & Grauert define a complex $r$-form at $x\in M$ as an alternating multilinear map 
$$
\omega\colon T_xM \times \dots \times T_xM \to \mathbb C.
$$
Any such form can be decomposed into its real and imaginary parts: $\omega = \alpha + i\beta$, where $\alpha$ and $\beta$ are ordinary real-valued $r$-forms. If you plug this into the definition of $\bar\omega $, you'll find that $\bar \omega = \alpha - i\beta.$
If $z  = u+iv$ is a complex-valued smooth function, then $dz$ is by definition the complex $1$-form defined by $dz = du + i \, dv$.  Thus 
$d\bar z = d (u-iv) = du - i\,dv = \overline{dz}$.
Now if $z^j=x^j+iv^j$ is one of the complex coordinate functions in a holomorphic coordinate chart $(z^1,\dots,z^n)$, then the complex vector fields $\partial/\partial z^j$ and $\partial/\partial \bar z^j$ are defined by
$$
\frac{\partial}{\partial z^j} = \frac{1}{2} \left( \frac{\partial}{\partial x^j} - i \frac{\partial} {\partial y^j}\right), \qquad
\frac{\partial}{\partial \bar z^j} = \frac{1}{2} \left( \frac{\partial}{\partial x^j} + i \frac{\partial} {\partial y^j}\right).
$$
Then $\overline{\partial /\partial z^j} = \partial /\partial \bar z^j$, and you can check that the definitions yield 
$$
dz^j \left( \frac{\partial}{\partial z^j} \right) =1 ,\qquad
dz^j \left( \frac{\partial}{\partial \bar z^j} \right) =0,
$$ 
along with the analogous formulas for $d\bar z^j$ obtained by conjugation:
$$
d\bar z^j \left( \frac{\partial}{\partial \bar z^j} \right) =1 ,\qquad
d\bar z^j \left( \frac{\partial}{\partial z^j} \right) =0.
$$ 
