Consider the map $F:\mathbb{R}^3 \to \mathbb{R}^2$ defined by $(x,y,z) \to (xz-y^2,yz-x^2)$. For which values of $(a,b) \in \mathbb{R}^2$ is $F^{-1}(a,b)$ a smooth manifold of $\mathbb{R}^3$?
$$DF=\begin{bmatrix} z & -2y & x \\ -2x & z & y \\ \end{bmatrix}$$
The matrix has full rank unless $y=x, z=2x $ or $y=x,z=-2x$. Since $F(x,x,2x)=(x^2,x^2)$ and $F(x,x,-2x)=(-3x^2,-3x^2)$, we have that $F^{-1}(a,b)$ is a smooth manifold for $a\ne b$. The only points to check now are of the form $(a,a)$. How do I check for these points??
Thanks for the help!!
Besides those $(a,b)$ which are regular values (in which case the regular value theorem applies as you have explained), I do not think there are any general methods. I can only suggest ad hoc methods to try.
For example, first identify those points in $F^{-1}(a,a)$ for which $DF$ is indeed of rank $2$; by application of the regular value theorem, you are guaranteed that $F^{-1}(a,a)$ is a smooth manifold in a neighborhood of each such point. Next write down the set of points in $F^{-1}(a,a)$ for which $DF$ has rank less than $2$. If you are lucky, it's a finite set, and you can then just examine $F^{-1}(a,a)$ locally near each such point and check one-by-one whether $F^{-1}(a,a)$ is a manifold near each such point. If you aren't lucky... well... I don't know what to say.