Suppose $f:\mathbb R^n\to \mathbb R^n$ is a $\mathcal C^1$ map such that $\langle df_x(v),v\rangle>0$ for all $x\in \mathbb R^n$ and $v\in R^n\setminus \{0\}$. Prove that $f$ is injective. Hint: For $x\ne 0, $ consider $g:\mathbb R\to \mathbb R^n$ given by $g(t)=f(tx)$. Find $g'(t)$ and show that $f(x)\ne f(0)$.
Let $x=(x_1,\dots,x_n)$. To compute $g'(t)=dg_t$, note that $g$ is the composition of $h:\mathbb R\to \mathbb R^n$ given by $t\mapsto tx$ and $f$. By the chain rule, $dg_t=df_{tx}\cdot dh_t$ as matrices. Now $dh_t=[x_1,\dots,x_n]^t,df_{tx}=[D_if_j]$, where $f=(f_1,\dots,f_n)$, so $$dg_t=[D_1f_1(tx)x_1+\dots+D_1f_n(tx)x_n,\dots, D_nf_1(tx)x_1+\dots +D_nf_n(tx)x_n]^t.$$
Injectivity means that $g(0)=f(x)\ne 0$ if $x\ne 0$. What does it have to do with injectivity? How to use the given inequality?
So $g=f \circ h$. $h$ is linear. Hence its derivative at each point is itself. Which means that for each $t$, $h^\prime(t)$ is the real map $h^\prime(t): u\mapsto ux$.
Consequently using the chain rule, we have:
$$g^\prime(t) =f^\prime(tx).x$$
Now consider $G(t)= \langle g(t), x \rangle$. $G$ derivative is $$G^\prime(t) = \langle g^\prime(t),x\rangle = \langle f^\prime(tx).x,x\rangle $$
which is strictly positive for $x \neq 0$ according to the hypothesis $\langle f^\prime(y).v,v\rangle >0$
Then $$\langle f(x)-f(0),x\rangle = G(1)-G(0)=\int _0^1 G^\prime(t) \ dt>0$$
allowing to conclude that $f(x)\neq f(0)$. The injectivity of $f$ can be deduced by changing $g(tx)$ into $g(t(y-x)+x)$.