The gauge transformation of the metric tensor is supposed to be $$\delta_\xi g_{\mu\nu}=\xi^\rho\partial_\rho g_{\mu\nu}+\partial_\mu\xi^\rho g_{\rho\nu}+\partial_\nu\xi^\rho g_{\mu\rho}$$ with $\delta_\xi x^\mu=-\xi^\mu(\tau)$, where $\xi^\mu$ is infinitesimal. What is the gauge transformation of the inverse metric $g^{\mu\nu}$?
My attempt:
We know $0=\delta(g^{-1}g)=\delta(g^{-1})g+g^{-1}\delta g\,$ and therefore $\delta(g^{-1})=-g^{-1}\delta (g) g^{-1}$
With $(g_{\mu\nu})^{-1}=g^{\mu\nu}$ I get: $$\begin{align} \delta_\xi g^{\mu\nu}&=-g^{\mu\nu}(\xi^\rho\partial_\rho g_{\mu\nu}+\partial_\mu\xi^\rho g_{\rho\nu}+\partial_\nu\xi^\rho g_{\mu\rho})g^{\mu\nu}\\ &=-g^{\mu\nu}(\xi^\rho(\partial_\rho g_{\mu\nu}) g^{\mu\nu}+\partial_\mu\xi^\rho g_{\rho\nu}g^{\mu\nu}+\partial_\nu\xi^\rho g_{\mu\rho}g^{\mu\nu})\\ &=-g^{\mu\nu}(\xi^\rho(\partial_\rho g_{\mu\nu}) g^{\mu\nu} + \partial_\mu\xi^\rho\delta_\rho^\mu +\partial_\nu\xi^\rho\delta_\rho^\nu)\\ &=-g^{\mu\nu}\xi^\rho(\partial_\rho g_{\mu\nu}) g^{\mu\nu}-g^{\nu}_\rho\partial_\mu\xi^\rho-g^{\mu}_\rho\partial_\nu\xi^\rho \end{align}$$
Is this the right way? The first term looks erroneous. And I am not quite sure if it is legal what I did with the deltas.