I have some doubts on a question related to the existence of a never vanishing smooth vector field on a manifold.
First of all I have proved that $M=F^{-1}(1)$ is a regular submanifold of $\mathbb{R}^4$ where $F$ is defined as: \begin{equation} F(x,y,z,t)=x^2+y^2+z^2-t^2. \end{equation} Then I proved that $M$ is diffeomorphic to $S^2\times\mathbb{R}$ using the following map; since for $(x,y,z,t)\in M$ we have $\sqrt{x^2+y^2+z^2}\neq 0$, for a fixed $t$ I can define: \begin{equation} F_t:S^2(\sqrt{1+t^2})\to S^2(1) \ \ \ F_t(x,y,z)=\bigg(\dfrac{x}{\sqrt{x^2+y^2+z^2}},\dfrac{y}{\sqrt{x^2+y^2+z^2}},\dfrac{z}{\sqrt{x^2+y^2+z^2}}\bigg). \end{equation} Then I simply notice that $F\colon M\to S^2\times\mathbb{R}$ defined as $F(x,y,z,t)=(F_t(x,y,z),t)$ is a diffeomorphism because both $F_t$ and the identity are diffeomorphism.
Finally I have the problematic question: does there exist a never vanishing vector field on $M$?
I know that to answer it, the Hairy ball theorem (telling us that on $n$-spheres with $n$ even, there is no never vanishing vector field) should be useful; but how can I use this statement?
You're almost done. Once you know that $M \simeq \Bbb S^2 \times \Bbb R$, let $f\colon M \to \Bbb S^2 \times \Bbb R$ be such a diffeomorphism. On the RHS, the vector field $\frac{\partial}{\partial t}$ (where $t$ is the $\Bbb R$ coordinate) is never vanishing, and so is $X = f^*\frac{\partial}{\partial t}$ on $M$!
Just a comment: the hairy ball Theorem tells you that you cannot count on the $\Bbb S^2$ component of $M$ to find a vector field that does not vanish. Hence, if such a vector field exists, it surely is thanks to the $\Bbb R$ component.