Let $f_i\colon \mathbb{R}^4\rightarrow \mathbb{R},\ i=1,2,3$ be defined by
$f_1(x)=x_1x_3-x_2^2 \\
f_2(x)=x_2x_4 -x_3^2\\
f_3(x)=x_1x_4-x_2x_3$
I want to show that $M=\{x\in\mathbb{R}^4 -\{0\} \colon f_1(x)=f_2(x)=f_3(x)=0\}$ is a 2-dimensional submanifold of $\mathbb{R}^4$.
My problem is, that I can only show that it is a 1 dimensional submanfold.
It is clear, that every $f_i\in\mathcal{C}^1(\mathbb{R}^4)$ and that the functional matrix
$\frac{\partial (f_1,f_2,f_3)}{\partial( x_1,x_2,x_3,x_4)}=
\begin{pmatrix}
x_3 & -2x_2 & x_1 & 0 \\
0 & x_4 & -2x_3 & x_2 \\
x_4 & -x_3 & -x_2 & x_1
\end{pmatrix}$
has rank 3, therefore it is surjective. This should show that $M$ is a 1 dimensional submanifold.
At this point I have no idea, since if I remember correctly, the dimension of a submanifold is unique (think it follows from the uniqueness of the atlas for $M$).
Best Answer
The set of points that satisfy the equations $f_1(x)=f_2(x)=f_3(x)=0$ can be described as the image of the map $$ \mathbb{R}^2\rightarrow\mathbb{R}^4, $$ where $$ (s,t)\mapsto(s^3,s^2t,st^2,t^3). $$ Plugging these points into the Jacobian matrix that you wrote above gives $$ \begin{bmatrix} st^2&-2s^2t&s^3&0\\ 0&t^3&-2st^2&s^2t\\ t^3&-st^2&-s^2t&s^3 \end{bmatrix} $$
Now, observe that $t$ times row $1$ minus $s$ times row $3$ is $$ \begin{bmatrix}st^3&-2s^2t^2&s^3t&0\end{bmatrix}-\begin{bmatrix}st^3&-s^2t^2&-s^3t&s^4\end{bmatrix}=\begin{bmatrix}0&-s^2t^2&2s^3t&-s^4\end{bmatrix}, $$ which is $-\frac{s^2}{t}$ times row $2$. So, the rank of this image is $2$. At all (most?) other points, the rank is $3$ (this is what is meant by generically).