How to make integral substitution from 3d k-space to 1d energy domain with non-trivial dispersion relation

In solid state phyics, one often encounters a subsitution from the 3d k space to a 1d integral via the standard dispersion relation $E= vec{k}^2/2m$ via

begin{align}
int dvec{k} Z(vec{k}) = int_0^infty dE mathcal{D}(E)
end{align}

where we have the two different density of states $Z$ and $mathcal{D}$. This is mainly solved by considering $dvec{k} Z(vec{k}) = dE mathcal{D}(E)$ where we use $dvec{k}=4pi k^2 dk$ and $frac{dE}{dk}=k/m$ and then use that there is (luckily) a mapping from $k=|vec{k}|$ to $E$. I never really questioned this although this method we used is not mathmatically rigorous especially if one encounters another type of dispersion.

In my case, the dispersion is given by $E=cos(k_x)+cos(k_y)$ and the procedure above fails because the dispersion is not rotationally symmetric. As the subsitution $(k_x,k_y) to E$ is not bijective, I fail to find a density of states $mathcal{D}(E)$.

I am asking for a mathematically rigorous proof of the case for the symmetric dispersion, so mainly non-bijective subsitution which can also be applied to the non-rotationally symmetric dispersion.