Look at the following equation system:
$3x+y-z+u^4=0 \\ x-y+2z+u=0 \\2x+2y-3z+u=0 $
Prove that the system doesn't fix $(x,y,z)$ as a function of $u\in(-\delta,\delta)$ for all $\delta>0$.
My question is If I prove that the assumptions of the implicit function theorem don't apply in this case, specifically that \begin{bmatrix} \overrightarrow{\frac{df}{dx} }&\overrightarrow{ \frac{df}{dy} }& \overrightarrow{\frac{df}{dz}} \end{bmatrix} isn't invertible for $f(x,y,u,z)=(3x+y-z+u^4,x-y+2z+u,2x+2y-3z+u)$ . Can we conclude from that that the system doesn't fix $(x,y,z)$ as a function of $u\in(-\delta,\delta)$ for all $\delta>0$?
If not, how can I solve this?
No, in general that's not enough. (The implicit function theorem gives a sufficient but not a necessary condition.)
But in this case, if you let $u=0$, the system becomes $$ \begin{pmatrix} 3 & 1 & -1 \\ 1 & -1 & 2 \\ 2 & 2 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} =\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} , $$ which has nontrivial solutions (since the determinant is zero), namely $(x,y,z)=(-t,7t,4t)$, $t \in \mathbf{R}$. So the system doesn't even define the values $x(0)$, $y(0)$, $z(0)$ uniquely.