The (geometric) Implicit function theorem states that:
If $f:U\subset\mathbb{R}^3\to\mathbb{R}$ is a differentiable function, and $a\in f(U)$ is a regular value of $f$, then $f^{-1}(\{a\})$ is a regular surface in $\mathbb{R}^3$.
Hence, the theorem provides us with a simple sufficient condition for a surface to be regular.
Is the condition also necessary? That is, does there exist a surface given by the equation $f(x,y,z)=0$, where $f$ is again differentiable but $0$ is not a regular value of $f$, such that the surface generated by the equation is regular?