In the question Calculating the determinant of the Hessian of a function, we know that the Hessian of function $$f(x_1,\dots,x_n) = g\bigg(x_1,\bigg(\sum_{i=2}^n x_i^2\bigg)^{1/2}\bigg), $$
and denote \begin{equation} t:=x_1 \quad \text{and}\quad r:= \bigg(\sum_{i=2}^n x_i^2\bigg)^{1/2}, \end{equation} is given by
\begin{align} D^2 f &= J_1^T\big(H_1 - \frac{g_r}{r}I_2\big)J_1 + \frac{g_r}{r}I_n , \end{align} where $I_k$ is the $k\times k$ identity matrix, $$J_1 = \left( \begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0 \\ 0 & \frac{x_2}{r} & \frac{x_3}{r} & \cdots & \frac{x_n}{r} \\ \end{array} \right), $$ and $$H_1 = \left( \begin{array}{cc} g_{tt} & g_{tr} \\ g_{rt} & g_{rr} \\ \end{array} \right). $$
Then the determinant of the Hessian of $f$ is given by \begin{equation} \det D^2f = (g_{tt}g_{rr}-g_{tr}^2)\bigg(\frac{g_r}{r}\bigg)^{n-2}. \end{equation}
My question is how to calculate the inverse of Hessian $ D^2 f$, or how to get the following formula $$\sum_{i,j=1}^nf^{ij}w_{ij}=(n-2)\frac{r}{g_r}\frac{w_r}{r}+\frac{g_{tt}w_{rr}+g_{rr}w_{tt}-2g_{rt}w_{rt}}{g_{rr}g_{tt}-g_{rt}^2},$$ where $\{f^{ij}\}$ is the inverse matrix of $ D^2 f$ and $w=(\det D^2 f)^{-\frac{n+1}{n+2}}.$
This question come from the paper The Bernstein problem for affine maximal hypersurfaces by Neil S. Trudinger and Xu-Jia Wang, I have made efforts to calculate it, but failed. I hope I can find some good ideas here, thank you.
Finally, I figured it out. I will use the well-known formula $$\left.\frac{d}{ds}\det(A+sB)\right|_{s=0}=\det A \cdot\text{tr}(A^{-1}B)\,.$$
We denote $D^2f=(f_{ij})$ and $D^2 w=(w_{ij})$, and note that $$\det D^2 f=\left(\frac{g_r}{r}\right)^{n-2}\left(g_{rr}g_{tt}-g_{rt}^2\right),$$ then we have $$\begin{aligned}\det\big((f_{ij})&+s(w_{ij})\big)\\&=\left(\frac{g_r+sw_r}{r}\right)^{n-2}\left((g_{rr}+sw_{rr})(g_{tt}+sw_{tt})-(g_{rt}+sw_{rt})^2\right),\end{aligned}$$ differentiating it at $s=0$ directly yields $$\begin{aligned}(n-2)\left(\frac{g_r}{r}\right)^{n-3}\frac{w_r}{r} \left(g_{rr}g_{tt}-g_{rt}^2\right) &+\left(\frac{g_r}{r}\right)^{n-2}(g_{rr}w_{tt}+g_{tt}w_{rr}-2g_{rt}w_{rt})\\&=\det D^2f\cdot\text{tr}\big((D^2f)^{-1}D^2w\big)\,.\end{aligned}$$ Hence, we obtain what we want.