Let be $h:\mathbb{R}^n\setminus\{0\}\to\mathbb{R}$ a continuous function,
$f:\mathbb{R}^n\setminus\{0\}\to\mathbb{R}^n$ a vector field where $f(x):=h(x)x$,
$C$ a continuously differentiable curve with $C\subset\{x\in\mathbb{R}^n\mid \Vert x\Vert_2=R\}$, where $R>0$ and $\int_C f(x)\cdot dx = 0.$
Show that if $f$ has an antiderivative $F$, then for all $y,z\in\mathbb{R}^n\setminus\{0\}$ with $\Vert y\Vert_2=\Vert z \Vert_2$ it holds
$F(y)=F(z)$ and $h(y)=h(z)$.
My approach:
My idea so far was to find a continuously differentiable parametrization $\varphi:[\alpha,\beta]\to\mathbb{R}^n$ of a curve $C$ that connects two points $y$ and $z$. We assume $\Vert y\Vert_2=\Vert z\Vert_2=R>0$ and define $$ \varphi[0,1]\to\mathbb{R}^n, \text{ where } \varphi(t)=\frac{R}{\Vert\left(y+t(z-y)\right)\Vert_2}\left(y+t(z-y)\right). $$ So for all $t\in[\alpha,\beta]$ we have $\Vert \varphi(t)\Vert_2=R$ and by the given assumption this means that $\int_C f(x)\cdot dx = 0$. As $F$ is an antiderivative we can conclude $$ 0=\int_C f(x)\cdot dx = F(y)-F(z)\implies F(y)=F(z). $$ However, it could be that $y+t(z-y)=0$. Maybe there is another "easy" parametrization that excludes this case? Or is this approach useless? Do you have any hints for the part $h(y)=h(z)$?
EDIT:
I made some progress regarding the part $h(y)=h(z)$:
We assume that we can find a curve $C$ with a continuously differentiable parametrization $\varphi:[\alpha,\beta]\to\mathbb{R}\setminus\{0\}$ with $\varphi(\alpha)=y$ and $\varphi(\beta)=z$ such that $\varphi([\alpha,\beta])\subset \{x\in\mathbb{R}^n\mid \Vert x\Vert_2=R\}$.
Then, $\int_C f(x)\cdot dx = F(y)-F(z)=0$ because $F$ is the antiderivative of $f$. So $h(y)y=h(z)z$.
If $h(z)=0$ then $h(y)=0$ because $y\neq0$. If $h(z)\neq0$ then $z=\frac{h(y)}{h(z)}y$ and from $\Vert y\Vert_2=\Vert z \Vert_2$ it follows immediately $h(y)=h(z)$.