Proof a set is a submanifold Let $M=\{x\in \mathbb{R}^3\colon f(x)=g(x)=0\}$ and 
$f\colon \mathbb{R}^3\longrightarrow \mathbb{R}, \ \ f(x_1,x_2,x_3)=-x_1^2-2x_1x_2+2x_2+2x_3\\
g\colon \mathbb{R}^3\longrightarrow \mathbb{R}, \ \ g(x_1,x_2,x_3)=\frac{3}{2}x_1^2-x_1x_2-3x_2+x_3$ 
be given.
I want to show that $M$ is a onedimensional submanifold of $\mathbb{R}^3$. 
The only reasonable approach from what I have learned so far, would be the regular value theorem. But to apply the theorem I would need an implicit function describing the manifold. Not sure if I miss something obvious.
 A: I imagine that what you call the "regular value theorem" is that if $\Phi : \mathbb{R}^n \to \mathbb{R}^m$ is smooth and $y \in \mathbb{R}^m$ is a regular value (i.e. for all $x \in f^{-1}(y)$, the differential $D_x\Phi$ is surjective) then $\Phi^{-1}(y)$ is a submanifold of dimension $n-m$.
A standard thing to do in your case is to consider the map $\Phi : \mathbb{R}^3 \to \mathbb{R}^2$, given by $\Phi(x) = (f(x), g(x))$. Then you're looking at $\Phi^{-1}(0)$ and you want to apply the regular value theorem to that.
So you're left with computing the differential $D_x\Phi$ at some point $x = (x_1,x_2,x_3) \in \Phi^{-1}(0)$. The matrix of the differential $D_x\Phi$ is given by the Jacobian of $\Phi$:
$$\operatorname{Jac}_{x}(\Phi) =
\begin{pmatrix}
-2x_1-x_2 & -2x_1+2 & 2 \\
3x_1-x_2 & -x_1-3 & 1
\end{pmatrix}$$
And what you now need to check is that if $\Phi(x) = 0$ then this matrix has full range, i.e. the linear map it represents is surjective. Using the rank-nullity theorem you may alternatively check that the null space of this matrix is of dimension $1$, whatever you prefer.
A: The implicit function that you need is already in the question that you wrote.
Let me describe it more precisely:
Let $F:\mathbb{R}^3 \rightarrow \mathbb{R}^2$ be a the implicit function defined as
$$
F(x_1,x_2,x_3) =(f(x_1,x_2,x_3),g(x_1,x_2,x_3))=(x_1^2-2x_1x_2+2x_2+2x_3, \frac{3}{2}x_1^2-x_1x_2-3x_2+x_3) 
$$
The Jacobian matrix of $F$ is,
$$
J_{F} = \left[ \begin{array}{cc}
     \frac{\partial f}{\partial x_1} & \frac{\partial f}{\partial x_2} & \frac{\partial f}{\partial x_3} \\
     \frac{\partial g}{\partial x_1} & \frac{\partial g}{\partial x_2} & \frac{\partial g}{\partial x_3} 
\end{array}\right]= \left[ \begin{array}{cc}
    -2x_1-2x_2 & -2x_1+2 & 2\\
    3x_1-x_2 & -x_1-3 & 1 
\end{array}\right]
$$
To prove that $M$ is submanifold of dimension $1$ is sufficient to prove that it exists a submatrix $2 \times 2$ of $J_F$ with a non-zero determinant. Thus
$$
\left| \begin{array}{cc}
     -2x_1+2 & 2\\
     -x_1-3 & 1 \end{array} \right| = 8.
$$
A: Apply the implicit function theorem to 
$$F=\left(\array{f \\ g}\right):\mathbb{R}^3 \rightarrow \mathbb{R}^2 $$
