Suppose some power series $f_1, \dotsc, f_k \in \mathcal O_0 = \mathbb C\{x_1, \dotsc, x_n\}$ define a holomorpic map $$f: \mathbb D^n_{\epsilon} \to \mathbb C^k,$$ where $\mathbb D^n_{\epsilon}$ denotes the polydisc of radius $\epsilon$ in $\mathbb C^n$. Is it possible to choose coordinates $y_1, \dotsc, y_{n-k}, z_1, \dotsc, z_k$ such that $$\frac{\partial f_i}{\partial y_j} = 0 \quad \forall i,j,$$ possibly after shrinking $\mathbb D^n_{\epsilon}$? By the theorem on implicit functions this works if the differential matrix $$Df(0) = \left(\frac{\partial f_i}{\partial x_j}(0)\right)_{ij}$$ has full rank $k$. I'm interested in the case where $Df(0)$ does not have full rank, but the reduced fiber $f^{-1}(0)_{red} \subset \mathbb D^n$ is still smooth. And of course I don't need the full strength of the implicit function theorem.
Here is an easy example: Consider $f: \mathbb C^2 \to \mathbb C, (x,y) \mapsto (x+y)^2$. If we introduce new coordinates $(x,z) = (x, x+y)$, then $f$ becomes $\tilde f(x,z) = z^2$, and now $$\frac{\partial \tilde f}{\partial x} = 0.$$