What does $d\bar z$ mean? What does $d\bar z$ mean?
For a manifold, given a local coordinate, $dx$ acts on tangent vectors and gives its corresponding components.
What does $d\bar z$  do?
The complex field is a one dimensional complex manifold, given the standard coordinate $z$, what does $d\bar z$ even mean?
I saw this notation in some books when defining a metric tensor:
$$\lambda dz d\bar z\ .$$
 A: There is always some disagreement on such notation, but you should probably think of $dz$ as acting on tangent vectors (assuming we're in one dimension) by $dz(v) = v\in\Bbb C$ and $d\bar z$ as acting by $d\bar z(v) = \bar v$.  (In general, in working with complex manifolds, one works with the complexified tangent bundle, which is spanned by the "holomorphic" and "anti-holomorphic" vector fields.)
EDIT: Thanks to Jack. As I said, there is always disagreement and confusion (even among those of us who've worked in complex geometry). On one hand, we want $(dz\otimes d\bar z)(v,v)$ to mean $dz(v)\overline{dz(v)}=|v|^2$. But differentiating $\bar z = x - iy$, $d\bar z = dx - i\,dy$, we have $\overline{dz(v)} = d\bar z(\bar v)$. Ultimately, we need to decide whether we're computing the inner product by $(dz\otimes d\bar z)(v,v)$ or by $(dz\otimes d\bar z)(v,\bar v)$.  The occasional excellent and well-respected textbook gets these things muddled, too. As I said, confusing. I should never have weighed in :P
A: $z$ is just the complex function on the complex plane and $\bar z$ is its conjugate. Writing in usual Euclidean coordinate, $z =x+ \sqrt{-1} y$ and $\bar z = x - \sqrt{-1} y$, so 
$$dz =dx+ \sqrt{-1} dy \text{  and  } d\bar z = dx - \sqrt{-1} dy\ .$$ 
If you put this into the metric, 
$$\lambda dzd\bar z = \lambda (dx dx + dydy)\ ,$$
which means that this is a very special expression: In general the first fundamental form has three unknown $E, F$ and $G$, but for this one you have $E=G = \lambda$ and $F=0$. Such a metric is called a conformal metric.  
