In general, this can not be done. For example, in dimension $2$ in coordinates $(x,y)$, let $$ G(x,y) = \left[\begin{matrix}x&y\\y&-x\end{matrix}\right]. $$ If $G$ could be diagonalized by a differentiable invertible matrix $A(x,y)$, i.e., if $$ A^T G A = \left[\begin{matrix}\lambda_1&0\\ 0&\lambda_2\end{matrix}\right] $$ where $\lambda_1$ and $\lambda_2$ were differentiable, then the $\lambda_i$ would have to vanish at $x=y=0$. Taking determinants yields $$ -(x^2+y^2)(\det A)^2 = \lambda_1\lambda_2\,. $$ Then, looking at the lowest order terms on each side (i.e., the terms of order $2$), you'd have $x^2+y^2$ written as the product of two linear factors in $x$ and $y$, which is impossible.
Similarly, you cannot achieve $$ G = A^T\left[\begin{matrix}\lambda_1&0\\ 0&\lambda_2\end{matrix}\right]A $$ for a differentiable $A$ and $\lambda_i$.