I want to show that the set $M=\{(x^2,y^2,z^2,xy,xz,yz): (x,y,z)\in S^2\}$ is a smooth manifold of dimension $2$ by explicitly constructing a smooth atlas. My book proceeds as follows (with paraphrasing and different notation):
Using the real projective plane $S^2/\{p,-p\}$, we would arrive at such a chart for $M$ as $$\Phi:\mathbb D^2\to M$$ $$(x,y)\mapsto (x^2,y^2,1-x^2-y^2, xy,xz,yz),$$ where $\mathbb D^2$ is the open unit disk.
This is clearly a smooth map. But its inverse (which equals $(m_1,m_2,...,m_6)\mapsto (\sqrt m_1, \sqrt m_2)$) is not smooth. So I think this does not really define a chart. Are there any missing points?
Note that the first four coordinates of the mapping give a $4$-to-$1$ mapping near the origin. We have to use the last two coordinates to get a $1$-to-$1$ mapping. In particular, the inverse function theorem tells us that $$\phi(x,y) = \big(x\sqrt{1-x^2-y^2},y\sqrt{1-x^2-y^2}\big) = (v,w)$$ has a smooth inverse in a neighborhood of the origin, even if we cannot write down the explicit formula. Then you get a local smooth inverse of your mapping by taking $\phi^{-1}\circ\pi$, where $\pi$ is projection onto the last two coordinates. Both these maps are smooth.