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The short-hand notation here is $1 = x_1 , 2 = x_2 ,... $and $\int_{1}=\int{dx_1},\int_{2}=\int{dx_2}.... $

In appendix A of this paper https://arxiv.org/abs/hep-th/9908172 it is said that the basic properties of derivatives with respect to an unconstrained tensor and its inverse are

$$[\frac{\delta }{\delta H_{12}},\frac{\delta }{\delta H_{34}}]=[\frac{\delta }{\delta H_{12}^{-1}},\frac{\delta }{\delta H_{34}^{-1}}]=0,\tag{A1}$$

and

$$\frac{\delta {H_{12}} }{\delta H_{34}}=\frac{\delta {H_{12}^{-1}} }{\delta H_{34}^{-1}}=\delta_{13}\delta_{42},\tag{A2}$$

and so we have

$$0=\frac{\delta }{\delta H_{34}} \delta_{12}=\frac{\delta }{\delta H_{34}} \int_5{H_{15}^{-1}}H_{52}=\int_5{\frac{\delta H_{15}^{-1} }{\delta H_{34}}}H_{52}+\int_5{H_{15}^{-1}\frac{\delta H_{52} }{\delta H_{34}}}=\int_5{\frac{\delta H_{15}^{-1} }{\delta H_{34}}H_{52}}+H_{13}\delta_{24}.\tag{A3}$$

From this how can we conclude that $$\frac{\delta H_{12}^{-1}}{\delta H_{34}}=-H_{13}^{-1}H_{42}^{-1}~?\tag{A4}$$

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After the insight of Qmechanic i think got it correct me if i am wong. we have

$\int_5{\frac{\delta H_{15}^{-1} }{\delta H_{34}}H_{52}}=-H_{13}^{-1}\delta_{24}$

since $\delta_{24}$ is symmetric we have

$\int_5{\frac{\delta H_{15}^{-1} }{\delta H_{34}}H_{52}}=-H_{13}^{-1}\delta_{42}$

Multiplying both side by $\int_2{H_{22'}}$ we obtain

$\int_5\int_2{\frac{\delta H_{15}^{-1} }{\delta H_{34}}H_{52}H_{22'}^{-1}}=-\int_2 H_{13}^{-1}\delta_{42}H_{22'}^{-1}$

or

$\int_5{\frac{\delta H_{15}^{-1} }{\delta H_{34}}\delta_{52'}}=- H_{13}^{-1}H_{42'}^{-1}$

that is

$\frac{\delta H_{12'}^{-1} }{\delta H_{34}}=H_{13}^{-1}H_{42'}^{-1}$

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Hint: Eq. (A4) follows from the rule $$\delta (H^{-1})=-H^{-1}~\delta H ~H^{-1},$$ where $H$ is an invertible matrix and where $\delta$ is some operation that satisfies Leibniz rule.

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