Existence of a coordinate system How can we formally show that a coordinate system $(x,y)$ exists or does not exist? For instance for some given coordinate system $(r,\phi,\theta)$ defined on the manifold $M =(1,\infty)\times\mathbb{S}^2$, does there exist a coordinate system $(s,\phi,\theta)$ for some tensor:
 $$h=ds⊗ds+g(s)^2(d\phi⊗d\phi+\cos^2\phi \space d\theta⊗d\theta)$$ where $g(s)$ is a positive smooth function. 
NOTE: There is some $2$-tensor given on $M$ by:
$$f=\frac{r}{r^3+r-2}dr \otimes dr +r^2d\phi \otimes d\phi + r^2\cos^2\phi \space d\theta\otimes d\theta$$
This is a problem I am stuck with for days already, just do not know the formal way of proving such coordinate systems exist, given some manifold $M.$ I would appreciate the help. 
 A: This answer assumes the interpretation of the question I proposed in the comments: given the tensor 
$$h=\frac{r}{r^3+r-2}dr \otimes dr +r^2d\phi \otimes d\phi + r^2\cos^2\phi \space d\theta\otimes d\theta$$
defined on $M=(1,\infty)\times \mathbb S^2$ where $\theta,\phi$ are the standard spherical coordinates, find a change of coordinate $s=s(r)$ and a smooth function $g$ so that $$h=ds \otimes ds+g(s)^2(d\phi \otimes d\phi+\cos^2\phi \space d\theta\otimes d\theta).$$
We can immediately read off $r^2 = g(s)^2;$ so the hard part is finding the change of coordinate. Since $ds = \frac{ds}{dr}dr,$ equating the two expressions for $h$ given above yields $$\frac r{r^3 + r - 2}dr \otimes dr = \frac{ds}{dr}dr \otimes \frac{ds}{dr}dr$$ and thus $$\frac{ds}{dr} = \pm \sqrt{\frac{r}{r^3 + r - 2}}.$$ Since $r^3+r-2>0$ for $r>1,$ the RHS here is real with constant sign; so we can integrate it to find a general solution $s(r)$ which is injective and thus a valid change of coordinate:
$$s(r') = s(1) \pm \int_1^{r'}\sqrt{\frac{r}{r^3 + r - 2}}\,dr. $$ This integral has no expression in terms of elementary functions, so I would just leave it written like this; if you try a CAS you can get some horrible expression in terms of elliptic integrals.
