Explicit computation of the Hodge codifferential Question I'm given a Laplacian $\Delta_n=-4y^2 \cdot \frac{\partial^2}{\partial\bar{z} \partial z} + 4 iny \cdot \frac{\partial}{\partial\bar{z}}$, and I want  it to be the Laplace operator associated to a Cauchy-Riemann operator $\bar\partial:\Omega^{0,0}(E) \rightarrow \Omega^{0,1}(E)$, which is a differential operator of the form $\bar\partial=(\partial_{\bar{z}} + \alpha(z))d\bar{z}$. How to find a suitable Cauchy-Riemann operator?
Here $E$ is a smooth vector bundle over a compact Riemann surface $X$, and $\alpha(z)$ is a smooth function.
Since the problem is dependent on the metric we have on $X$, we have to specify it. Let $X$ be the quotient of the upper half plane $\mathbb{H}$ by a cocompact Fuchsian subgroup of the first kind, and the metric is the quotient of the hyperbolic metric on $\mathbb{H}$ by the action of the group. 

How to associate a Generalized Laplacian to a Cauchy-Riemann operator
The first step is to define an $L^2$-scalar product in
  $\Omega^{0,0}(E)$ and $\Omega^{0,1}(E)$. The hyperbolic metric on $X$
  corresponds to an Hermitian metric $h$ on the holomorphic tangent
  space of $X$ , i.e. the subbundle $T^{1,0}(X)$ of the complex tangent
  bundle of $X$. This metric induces a metric on the dual of
  $T^{1,0}(X)$ , i.e. differential forms of type $(1,0)$, and, by
  complex conjugation, on forms of type $(0,1)$. By taking the exterior
  powers of this metric, and by tensoring with the Hermitian metric on
  $E$ we get a pointwise scalar Hermitian product $(s(x),t(x))$ for two
  sections of $\Omega^{0,i}(E)=\Omega^{0,i}(X) \otimes_{C^\infty(X)}
 \Gamma(X,E)$. On the other hand, let $\omega_0$ be the normalized
  Kähler form: 
$\omega_0=\frac{i}{2\pi}h(\frac{\partial}{\partial z},
\frac{\partial}{\partial z}) dz d\bar{z}$
Then the $L^2$ scalar product of two sections $s,t \in
\Omega^{0,i}(E)$ is defined by:
  $(s,t)_{L^2}=\int_X(s(x),t(x))\omega_0$.
The second step consists in defining the operator $\bar{\partial}^*$
  in such a way that it is the $L^2$-adjoint of $\bar{\partial}$. So
  $\bar\partial^*:\Omega^{0,1}(E) \rightarrow \Omega^{0,0}(E)$ and
  $(s,\bar\partial^* t)_{L^2}=(\bar\partial s, t)_{L^2}$. Then the
  Laplace operator is defined by:
  $\Delta_n=\bar\partial^*\bar\partial:\Omega^{0,0}(E) \rightarrow
\Omega^{0,0}(E)$.
  
  
*
  
*This blockquote is a, more or less faithfully, quote from
  "Lectures on Arakelov Geometry" of Soulé, Abramovich, Burnol, Kramer.
  Chapter V, §2.2.*
  

My Approach My idea to find the right Cauchy-Riemann operator is to compute in full generality it's $L^2$-adjoint and find a general expression for the induced Laplacian. 
My Computations As Willie kindly pointed out I was working on the wrong definition of Laplacian, so all my previous computations were useless. I will post some progress here as soon as I will have them.
Thank you very much!
 A: Some obvious problems:
$*dy = - dx$ so $*(dz) = *(dx + i dy) = dy - i dx = -i dz$. This is actually what you expect, since $dz$ is null and is supposed to be either self-dual or anti-self-dual. (Also, on a general complex manifold one has the the Hodge star operator take $\Omega^{p,q}$ to $\Omega^{n-q,n-p}$ where $n$ is the complex dimension. You really should not have $dz$ transformed into $d\bar{z}$. 
Secondly, when you apply $df$, you should get $\left( \partial_{\bar{z}}f(z,\bar{z}) + \alpha(z)f(z,\bar{z})\right) \mathrm{d}\bar{z}$ if I understand your notation right. 
Thirdly, if you write $f = f(z)$, it is holomorphic, and thus $\partial_{\bar{z}} f = 0$. You should really be carrying around both $z$ and $\bar{z}$ coordinates. 
Lastly, you should not be trying to compute the Hodge Laplacian, at least not from what you've started with. What you wrote as $d$ is probably more properly written as $\bar{\partial}^\alpha$ (or something like that). It is not the exterior differential, which is $d = \partial + \bar{\partial}$. By definition $*\bar{\partial}f = i \bar{\partial}f$, so $\bar{\partial}*\bar{\partial}f = 0$ using that $\bar{\partial}^2 = 0$. 
I think you may want to double check your definitions to make sure you are asking the question you want to ask. Or perhaps you can ask a new question giving full details of your motivation (with preferably linked references) so we can address the above issues. In particular, it would be nice if you specified what it means to be "the Laplacian associated to a Cauchy-Riemann operator". 
A: A suitable Cauchy-Riemann operator is: $\bar \partial= 2 \partial_{\bar z}$. Indeed if the metric on $X$ is given by $h(s(z)dz, t(z)dz)= s(z) \overline{t(z)}\rho (z)$ the its $L^2$ dual  is $\bar \partial ^*=-\rho(z)^{1-n} 2\partial_z \rho(z)^n$.
Using the hyperbolic metric, i.e. $\rho(z)= y^2$, and taking the product $\bar \partial ^*\cdot \bar \partial $ we have the Laplacian as in the question.
This answer has been posted in such a way to have an explicit formula associated to the question, because the accepted answer is very useful but not a complete answer.
