I was reading the proof here for the inverse function theorem. I understand everything except in the middle of page 2, the proof says
$$\sum^n_{i=1}-2(f_i(x_0) - y_i) \frac{\partial{f_i}}{\partial{x_j}}(x_0) = 0$$
Since we know that $det(Df(x_0))$ is not zero and thus invertible, $f_i(x_0) - y_i = 0$.
This is used to show that $f$ maps an open set to an open set. My question: since $Df(x_0)$ is invertible so it only maps from $0$ to $0$. Then what prevents $x_0$ from being $0$?
Regard ${z_i}:=-2(f_i(x_0)-y_i)$ and let $\vec{z}=(z_1,z_2.\cdots,z_n)$ as a vector. Now, that expression above is the matrix $Df(x_0)$ multiplied into $z$. Since this product is zero and derivative matrix is nonsingular, we get $z=0$.