I am late, but I'll answer anyway. In linearized gravity, $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$. We wish to find the expression for $g^{\mu \nu}$ up to linear order, which in general, must be some linear combination of $\eta^{\mu \nu}$ and $h^{\mu \nu}$. To that end, we write:

$g^{\mu \nu} = a\eta^{\mu \nu} + b h^{\mu \nu}$ where $a,b$ are coefficients to be determined

$g_{\mu \nu} g^{\nu \lambda} = (\eta_{\mu \nu} + h_{\mu \nu})(a\eta^{\nu \lambda} + b h^{\nu \lambda}) = \delta_\mu^\lambda$

$\Rightarrow a \delta_\mu^\lambda + (a+b)h_\mu^{\ \lambda} = \delta_\mu^\lambda \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ [$O(h^2)$ term neglected]

$\Rightarrow a=1, b=-1$

So the inverse metric tensor in linearized gravity turns out to be $g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu}$.

I am late, but I'll answer anyway. In linearized gravity, $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$. We wish to find the expression for $g^{\mu \nu}$ up to linear order, which in general, must be some linear combination of $\eta^{\mu \nu}$ and $h^{\mu \nu}$. To that end, we write:

$g^{\mu \nu} = a\eta^{\mu \nu} + b h^{\mu \nu}$ where $a,b$ are coefficients to be determined

$g_{\mu \nu} g^{\nu \lambda} = (\eta_{\mu \nu} + h_{\mu \nu})(a\eta^{\nu \lambda} + b h^{\nu \lambda}) = \delta_\mu^\lambda$

$\Rightarrow a \delta_\mu^\lambda + (a+b)h_\mu^{\ \lambda} = \delta_\mu^\lambda \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ [$O(h^2)$ term neglected in the previous step]

$\Rightarrow a=1, b=-1$

So the inverse metric tensor in linearized gravity turns out to be $g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu}$.