Values of $t \in \mathbb{R}$ for which $f^{-1}(t)$ is an embedded submanifold Let $f:\mathbb{R^3} \to \mathbb{R}$ be defined by $f(x,y,z)=(x-1)^2-yz$. For what values of $t \in \mathbb{R}$, is $f^{-1}(t)$ an embedded submanifold of $\mathbb{R^3}$ of dimension $2$?
The Jacobian of this function is of the form $[2x-2,-z,-y]$. The jacobian has full rank unless $x=1,y=0$ and $z=0$. Also $f(1,0,0)=0$. Thus I have to look at $f^{-1}(t)$ for $t=0$. Now the hessian matrix at $t=0$ is of the form $$\begin{bmatrix} 2 & 0 & 0\\ 0 & 0 & -1\\ 0 & -1 & 0\\ \end{bmatrix}$$
Let $D_1=2, D_2=\begin{bmatrix} 2 & 0 \\ 0 & 0 \\ \end{bmatrix},D_3=\begin{bmatrix} 2 & 0 & 0\\ 0 & 0 & -1\\ 0 & -1 & 0\\ \end{bmatrix}$. Then $|D_1|=2, |D_2|=0,|D_3| \ne 0$. Hence $(1,0,0)$ is a saddle point. Thus $f^{-1}(t)$ is an embedded submanifold except for when $t=0$. 
Is this alright?
Thanks for the help!!

In general, it might be difficult to tell if a set is an embedded smooth submanifold or not. However, in your case, you are lucky because $f^{-1}(0)$ is not even a two-dimensional topological manifold near $(1,0,0)$. 
Note that your function $f$ is a translated quadratic form (it is of the form $x'^2 = yz$ for $x' = x - 1$). This form has signature $(+,+,-)$ so the zero level set $S = f^{-1}(0)$ is an elliptic double cone (plot done using wolfram cloud):

Intuitively, this is not a topological two-dimensional manifold near the vertex $v = (1,0,0)$. To see this formally, note that any open neighborhood $U$ of $v$ in $S$ has the property that $U \setminus \{ v \}$ is disconnected (by throwing away $v$ we split the cone into two parts). Hence, $U$ cannot be homeomorphic to a two-dimensional open ball because if you throw away a point from a two-dimensional open ball, you are left with a connected space.
