Tian’s book Canonical metrics in Kähler geometry, page 52-53 says:
Assume that we have normal coordinates at the given point, so $g_{i\bar{j}}=\delta_{ij}$ and that the first order derivatives of $g$ vanish.
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which we can rewrite after substituting $$\frac{\partial^2g_{i\bar{j}}}{\partial z^k\partial \bar{z}^l}=-R_{i\bar{j}k\bar{l}} \text{ and } \frac{\partial^2g^{i\bar{j}}}{\partial z^k\partial \bar{z}^l}=R_{i\bar{j}k\bar{l}} $$ as…
I don’t understand the second equation, here is my computation:
$\frac{\partial^2g^{i\bar{j}}}{\partial z^k\partial \bar{z}^l}= \frac{\partial}{\partial\bar{z}^l}(\frac{\partial g^{i\bar{j}}}{\partial z^k})= \frac{\partial}{\partial\bar{z}^l} (-g^{i\bar{p}}\frac{\partial g_{q\bar{p}}}{\partial z^k}g^{q\bar{j}})= -g^{i\bar{p}}\frac{\partial^2 g_{q\bar{p}}}{\partial z^k\partial \bar{z}^l}g^{q\bar{j}}=-\frac{\partial^2 g_{j\bar{i}}}{\partial z^k\partial \bar{z}^l}=R_{j\bar{i}k\bar{l}}$.
Here $R_{i\bar{j}k\bar{l}}:= -\frac{\partial^2g_{i\bar{j}}}{\partial z^k\partial \bar{z}^l}+g^{p\bar{q}}\frac{\partial g_{p\bar{j}}}{\partial\bar{z}^l}\frac{\partial g_{i\bar{q}}}{\partial z^k}$, since I choose normal coordinate, the first order terms vanish.
Thanks.