I'm not sure how to proof this theorem.
Let $w=\sum_{i=1}^{n}f_idx_i$ on C conic set. If $f_i$ is homogeneus (p-degree) and $w$ is closed.
$\Rightarrow$ $w$ is exact and $$g=\frac{1}{p+1}\sum_{i=1}^{n}x_if_i$$ is a primitive.
I have thought about using Euler's theorem about Homogeneus function but I'm stuck.
Euler's Theorem:
f on conic set is homogeneus p degree $\Leftrightarrow$ pf(x)=$\nabla f(x)\cdot x$
My thoughts:
$$ (p+1)\frac{\partial}{\partial k}g(x)=f_k(x)+\sum_{i=1}^{n}x_i\frac{\partial}{\partial k}f_i=(p+1)f_k(x) $$
So $$ \frac{\partial}{\partial k}g(x)=f_k(x) $$ So, for each k, I found a potential $\nabla g$ which is equal to $\vec{f}$ and so $w$ is exact.
Is it correct?
There is a standard way to get a primitive on a star-shaped region by doing the line integral along rays from the star-point (in this case, the origin). In particular, doing the line integral along the line segment from the origin to $x$, we will have \begin{align*} g(x) = \int_0^x \omega &= \int_0^1 \sum f_i(tx)x_i\,dt = \int_0^1 \sum t^p f_i(x)x_i\,dt = \sum x_if_i(x) \int_0^1 t^p\,dt \\ & = \frac1{p+1}\sum x_if_i(x), \end{align*} as you desired. You can check that $dg = \omega$ because $\omega$ was closed to start with.
P.S. I don't think that Euler's Theorem on homogeneous functions is relevant.