Let $f:U \rightarrow\mathbb{R}^{m}$ continue in a convex set $U \subset \mathbb{R}^{m}$. If $\langle{f'(x)(v),v}\rangle>0$ for all $x \in U$ and for all $0 \neq v \in \mathbb{R}^{m}$, then $f$ is a diffeomorphism of $u$ on its image.
So first, i wanna show that f'(x) is an isomorphism, particulary i have to show f'(x) os bijective, for inyectivity i have to see that the kernel of the linear transformation is zero, as f is linear transformation its derivative is the same, but i don't understant clearly the relationship between the kernel and the inner product.
The next step is proof the surjectivity but with the dimensional theorem is clear, and when f'(x) is isomorphism i can use the inverse function theorem to proof f is diffeomorphism, taking into account that f is a linear transformation in particular is $c^{1}$ class.