Standard Hermitian metric for $\mathbb C$ In the middle of Page 42 of Ballmann's book,  the author defines the Hermitian metric by $h=g+i\omega$, where $g$ is the compatible Riemannian metric which is a Riemannian metric $g$ satisfying $g(X,Y)=g(JX,JY),\forall X,Y\in T_{\mathbb C}M$ and $\omega:=g(JX,Y)$ is the associated fundamental form of $g$.
Then in the next Page (p.43), 4.10 Examples. 1), the author says for the complex manifold $\mathbb C$, the Euclidean metric is $g=\frac{1}{2}dz\odot d\bar z=\frac{1}{2}(dz\otimes d\bar z+d\bar z\otimes dz)$ and $\omega=\frac{i}{2}dz\wedge d\bar z=\frac{i}{2}(dz\otimes d\bar z-d\bar z\otimes dz)$.
In this case, the Hermitian metric $h=g+i\omega=\frac{1}{2}(dz\otimes d\bar z+d\bar z\otimes dz)-\frac{1}{2}(dz\otimes d\bar z-d\bar z\otimes dz)=d\bar z\otimes dz$, does it mean the standard Hermitian metric of $\mathbb C$ is $d\bar z\otimes dz$?
Then how to evaluate the length of $\frac{\partial }{\partial z}$, or how to compute $|\frac{\partial }{\partial z}|^2$?
If $|\frac{\partial }{\partial z}|^2$ is defined by $h(\frac{\partial }{\partial z},\frac{\partial }{\partial \bar z})$, does it mean the length is zero?
By the way, how to compute $|\frac{\partial }{\partial \bar z}|^2$?
 A: Note Ballmann's convention for Hermitian metrics (quoting from page $1$):

As a rule we assume that Hermitian metrics on a given bundle $E$ are
conjugate linear in the first variable and complex linear in the
second.

With this convention, your calculation is accurate. If one wishes to use the convention that Hermitian metrics are complex-linear in the first variable instead, then the standard Hermitian product on $\mathbb{C}^n$ is $\mathrm{d}z\otimes\mathrm{d}\bar{z}=g-\mathrm{i}\,\omega$.
So for now we will use this definition: a Hermitian metric on a complex vector space $V$ satisfies $h(v_1,v_2)=\overline{h(v_2,v_1)}$ and $h(\lambda\,v_1,\mu\,v_2)=\bar{\lambda}\mu\,h(v_1,v_2)$. Now, from any complex vector space $V$ you can get a real vector space $V_{\mathbb{R}}$ which as a set is still $V$, but with scalar multiplication restricted to the real numbers. The complex multiplication on $V$ induces an endomorphism $J$ of $V_{\mathbb{R}}$ satisfying $J^2=-\mathrm{Id}$. From $V_{\mathbb{R}}$ we can define a new complex vector space, $V_{\mathbb{C}}:=V_{\mathbb{R}}\otimes\mathbb{C}$, and we extend $J$ complex-linearly to obtain an endomorphism of $V_{\mathbb{C}}$.
We can split $V_{\mathbb{C}}$ as a direct sum of the eigenspaces for $J$ corresponding to the eigenvalues $\pm\,\mathrm{i}$, obtaining $V_{\mathbb{C}}=V^{1,0}\oplus V^{0,1}$. Now, what is the relation between the original $V$ and these new spaces? You can check that, as complex vector spaces, $V\cong V^{1,0}$. The isomorphism is given by the map $$v\mapsto \frac{1}{2}\left(v-\mathrm{i}\,J\,v\right),$$ with inverse $$v\mapsto\mathrm{Re}(v)=\frac{1}{2}(v+\bar{v})\in V_{\mathbb{R}}.$$
Let's get to the case $V=\mathbb{C}^n$. With the usual notation, $\partial_{z}$ is a section of $V^{1,0}$, while $\partial_{\bar{z}}$ is in $V^{0,1}$. On the face of it, it does not make much sense to plug $\partial_{z}$ in the Hermitian product $h$, which is defined on $V$. The notation $\lvert\partial_{z}\rvert^2$ usually stands for $$\lvert\partial_{z}\rvert^2=g(\partial_{z},\partial_{\bar{z}})=\lvert\partial_{\bar{z}}\rvert^2.$$ It is easy to check that $g(\partial_{z},\partial_{z})=g(\partial_{\bar{z}},\partial_{\bar{z}})=0$.
So, in which sense $h=\mathrm{d}\bar{z}\otimes\mathrm{d}z$? This identity holds on the extended vector space $V_{\mathbb{C}}$, on which however $h$ is not a Hermitian product. Anyway, we can relate the Hermitian product on $V$ with the pairing on $V_{\mathbb{C}}$: $g(\partial_z,\partial_{\bar{z}})=\frac{1}{2}\overline{h(\partial_x,\partial_x)}=\frac{1}{2}h(\partial_x,\partial_x)$ (see Remark $4.9$ in Ballmann's book), so the "length" is positive as you'd expect.
Now, let's look at some computations. The expression for $h$ on $V_{\mathbb{C}}$ in coordinates shows that $h(\partial_z,\partial_{\bar{z}})=0$ and $h(\partial_{\bar{z}},\partial_z)=1$, but we can also check this without using the differentials $\mathrm{d}z$ and $\mathrm{d}\bar{z}$. Take as $\mathbb{C}$-basis for $\mathbb{C}$ the vector $\partial_x$ (where $x$ is the usual real coordinate), and let $\partial_y=\mathrm{i}\,\partial_x=:J\partial_x$. Then $\partial_x,\partial_y$ is an $\mathbb{R}$-basis for $V_{\mathbb{R}}$, and $\partial_z=\frac{1}{2}(\partial_x-\mathrm{i}\,\partial_y)$ is a $\mathbb{C}$-basis for $V^{1,0}$. Together with $\partial_{\bar{z}}$, it forms a $\mathbb{C}$-basis for $V_{\mathbb{C}}$. If we extend $g$ and $\omega$ to $V_{\mathbb{C}}$ complex-linearly, also $h$ is extended, so it makes sense to compute $h(\partial_z,\partial_{\bar{z}})$. As $h$ is now complex-linear in $V_{\mathbb{C}}$, we get $$h(\partial_z,\partial_{\bar{z}})=\frac{1}{4}h\left(\partial_x-\mathrm{i}\,\partial_y,\partial_x+\mathrm{i}\,\partial_y\right)=\frac{1}{4}\left(h(\partial_x,\partial_x)-\mathrm{i}\,h(\partial_y,\partial_x)+\mathrm{i}\,h(\partial_x,\partial_y)+h(\partial_y,\partial_y)\right).$$
Now, $\partial_x$ and $\partial_y=J\partial_x$ are elements of $V_{\mathbb{R}}=V$ (as sets). Since $J$ on $V_{\mathbb{R}}$ is the multiplication by $\mathrm{i}$ coming from $V$, using that $h$ is Hermitian in $V$ we get $$h(\partial_z,\partial_{\bar{z}})=\frac{1}{4}\left(h(\partial_x,\partial_x)-\mathrm{i}\,h(\mathrm{i}\partial_x,\partial_x)+\mathrm{i}\,h(\partial_x,\mathrm{i}\partial_x)+h(\mathrm{i}\partial_x,\mathrm{i}\partial_x)\right)=\frac{1}{4}\left(h(\partial_x,\partial_x)+(-\mathrm{i})^2\,h(\partial_x,\partial_x)+\mathrm{i}^2\,h(\partial_x,\partial_x)+(-\mathrm{i})\mathrm{i}\,h(\partial_x,\partial_x)\right)=0.$$
The analogous computation with $\partial_z$ and $\partial_{\bar{z}}$ switched will give you $1$.
