$0\in \mathbb{R}$ is not a regular value of g(f)
Solution 1:
Your gradient is not correct, but otherwise, I think you right. $0$ is indeed a regular value. You know that $g(f(x,y,z)) = y$. The gradient is $\nabla (g \circ f)(x,y,z) = (0, 1, 0)$, which has rank 1. In particular, for any $u \in \mathbb{R}$, we have that $\langle \nabla (g \circ f)(x,y,z), (0,u,0)\rangle = u$, showing that the derivative is surjective at each point.
PS: The LaTex code for composition is \circ