If $f$ is continuously partially differentiable with open non-empty domain $D\subset \mathbb{R}^N$, s.t. the determinant of its jacobian is non-zero for all $x\in D$, show the mapping: $D\rightarrow \mathbb{R}, \ \ \ x\mapsto \|f(x)\|$ has no local maximum.
Likewise, if $D$ is now compact, and $f$ has a continuous extension $g:\overline{D}\rightarrow \mathbb{R}^N$. Show the continuous map: $\overline D\rightarrow \mathbb{R}, \ \ \ x\mapsto \|g(x)\|$ has a maximum on $\partial D$. (Boundary).
I know if $D\subset \mathbb{R}^N$ is non empty, and $x_0$ is an interior point of $D$, then the function $f:D\rightarrow \mathbb{R}$ has a local maximum at $x_0$, if there is $\epsilon >0$, s.t. $B_{\epsilon}(x_0)\subset D$ and $f(x)\leq f(x_0)$ for all $x\in B_{\epsilon}(x_0)$.
I tried making use of the fact that because $f$ is continuously partially differentiable with open domain $D$, such that the determinant of its jacobian is non-zero for all $x\in D$, implies there is a neighborhood $V\subset D$ of $x_0\in D$, and $C>0$, s.t. $\|f(x)-f(x_0)\|\geq C\|x-x_0\|$. Yet I can't show $\|f(x)\|\geq \|f(x_0)\|$, which would imply there doesn't exist a local maximum.
Another thought for the first question, may be to note that because the function is locally injective for all $x\in D$, then $f$ is injective. Thus I have to prove somehow, that injective multivariable functions don't attain a local maximum.
I am not sure if this is the right way to approach the question, and any help would be much appreciated. Thanks!