Let $F:\mathbb{R}^4\to\mathbb{R}^2$ by $F_1(x,y,z,t)=x^2+y^2z^2-2zt^3$ and let $F_2(x,y,z,t)=2xy^{-1}-3z^2t^{-1}$. Show that $F_1=6$ and $F_2=-5$ is an everywhere smooth surface. Also, find the linearization of this surface around the point $(2,-2,1,1)$.
This seems like an implicit function theorem problem, but I'm not sure. Is the point to show that for any point $(x,y,z,t)$, the determinant $det\begin{bmatrix}\frac{\partial F_1}{\partial x} & \frac{\partial F_1}{\partial y}\\ \frac{\partial F_2}{\partial x} & \frac{\partial F_2}{\partial y}\end{bmatrix}$ is nonzero?