I am having some difficulties in understanding the proof given by Do Carmo of the following proposition.
I will run through the proof and comment on it as it progresses and place my questions accordingly.
This makes sense, as $a$ regular implies the differential $df_p$ is nonzero at each $p\in f^{-1}(a)$. I assume that what Do Carmo is trying to say in an uneconomical way is $F(x,y,z)=(u,v,t)$. The differential calculation and result is clear to me.
We can apply the inverse function theorem to $F$ since $dF_p$ has inverse. I can also see why the coordinate functions of $F^{-1}$ are differentiable.
I am not seeing at all how $z=g(u,v,a)$. We established that the third coordinate function of $F^{-1}$ is $z=g(u,v,t)$, where $u,v,t\in W$. There is no reason to believe every point in $W$ has $a$ as its third coordinate.
$F(x,y,z)=(x,y,f(x,y,z))=(u,v,t)$. $F^{-1}(u,v,t)=(u,v,g(u,v,t))$. If $(x,y,z)\in f^{-1}(a)$, $F^{-1}(F(x,y,z))=F^{-1}(x,y,f(x,y,z))=F^{-1}(x,y,a)=F^{-1}(u,v,a)=(u,v,g(u,v,a))=(x,y,z)$ implies that $z=g(u,v,a)$.