Assume that the algebraically independent polynomials $f_1,\ldots, f_n\in\mathbb{C}[x_1,\ldots, x_n]$ are such that the Jacobian matrix $\text{Jac}_{x_1,\ldots, x_n}^{f_1,\ldots, f_n}\in\mathbb{C}\setminus\{0\}$.
Is it true that every derivation $D$ of algebra $\mathbb{C}[f_1,\ldots, f_n]$ can be continued to the derivation $U$ of $\mathbb{C}[x_1,\ldots, x_n]$, in other words does there exists a derivation $U$ such that $U|_{\mathbb{C}[f_1,\ldots, f_n]}= D$?
Let $R=\mathbb{C}[f_1,\ldots,f_n]\subset\mathbb{C}[x_1,\ldots, x_n]=A$. A derivation $U$ of $A$ is completely determined by $U(x_i)$, since, then $U(P(x_1,\ldots,x_n))=\sum\frac{\partial P}{\partial x_i} U(x_i)$ for any $P\in A$. In particular, one has $U(f_i)=\sum \frac{\partial f_i}{\partial x_j}U(x_j)$. If $U_{|R}=D$, then $U(f_i)=D(f_i)$. Now using the invertibility of the Jacobian, one ca calculate $U(x_i)$ in terms of $D(f_i)$ and $\frac{\partial f_j}{\partial x_i}$ and then you have such a $U$ by the first sentence.
