Let $U \subset \mathbb{R^m}$ be an open subset. Let $f:U \to \mathbb{R^n}$ be a $C^1$-smooth map and suppose that $f(x)=0$ for some $x \in U$ and that $Df(x)$ is invertible(derivative map of f at x). Let $B$ be a ball around $x$ such that $Df(y)$ is invertible for $y \in B$ (the inverse function theorem(applicable for f at x), guarantees the existence of such a ball $B$). Define $N:B \to \mathbb{R^n}$ by $$N(y):=y-[Df(y)]^{-1}f(y)$$.Then show that we can find $r>0$ such that $\forall y \in B$ with $||y-x||<r$, the sequence $\{y,N(y),N(N(y)),\ldots \}$ converges to $x$.
We have $N(x)=x$, so my first thought was to try and show that the map $N$ is a contraction, so that the Banach Fixed Point Theorem would guarantee the existence of a unique limit for the sequence(which would precisely be the fixed point $x$).
$$||N(y)-N(x)|| \leq ||y-x||+||[Df(y)]^{-1}f(y)-0||$$
$$\leq ||y-x|| + ||Df(y)^{-1}||_{op}||f(y)||$$ (op:-operator norm) but I couldn't find a suitable bound for the second term in the last inequality. Furthermore, I couldn't find a candidate value of $r$ such that $N$ maps $\overline{B(x,r)}$ to $\overline{B(x,r)}$ (to ensure completeness criteria).
I managed to show that $DN(x)=0$,
$N(x+h)-N(x)=h-[Df(x+h)^{-1}]f(x+h)$. Let $f(x+h)=0+Df(x)h+E(h)$ where $\frac{E(h)}{||h||} \to 0$ as $h \to 0 \implies N(x+h)-N(x)=[I-Df(x+h)^{-1}Df(x)]h+E(h)$. The map $x \to Df(x)^{-1}$ is continuous, so we can write $Df(x+h)^{-1} = Df(x)^{-1} +E_1(h)$ and $E_1(h) \to 0$ as $h \to 0$. This shows that $DN(x)=0$.
but I couldn't progress any further.