Consider the set of $2n-1$ functions of $2n$ variables \begin{gather*} x_1y_1, x_2 y_2, \ldots, x_n y_n,\\ x_2 y_1^2,x_3 y_1^3, \ldots,x_n y_1^n. \end{gather*} I need prove that these functions are functionally independent. A long way is to find the rank of the jacobian $$ \begin{vmatrix} y_{{1}}&0&0&0&\ldots&x_{{1}}&0&0&0\\ 0&y_{{2}}&0&0&\ldots&0&x_{{2}}&0&0\\ 0&0&y_{{3}}&0&\ldots&0&0&x_{{3}}&0\\ \ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\ 0&0&\ldots&y_{{n}}&\ldots&0&0&\ldots&x_{{n}}\\ 0&{y_{{1}}}^{2}&0&0&\ldots&2x_{{2}}{y_{{1}}}&0&\ldots&0\\ 0&0&{y_{{1}}}^{3}&0&\ldots&3x_{{3}}{y_{{1}}}^{2}&0&\ldots&0\\ \ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots&\ldots &\ldots\\ 0&0&\ldots&{y_{{1}}}^{n}&\ldots&nx_{{n}}{y_{{1}}}^{n-1}&0&\ldots&0 \end{vmatrix} $$
but on this way I have got big and bad determinants.
Is there any other way to do it?
Here is some divide-and-conquer approach. First, the functions $x_2y_2, \dots , x_ny_n$ are the only terms that contain $y_i$, $i\ge 2$. Hence this set of functions is linearly independent and not contained in the linear hull of all the other functions.
So it remains to consider the functions $x_1y_1 \dots x_n y_1^n$. But all these are linearly independent because exactly one term depends on $x_i$, $i=1\dots n$.
Hence, the functions in question are linearly independent.