This is an exercise that I have been assigned for homework. I don't really know how to approach it though. I know that $\int_{\phi}\omega$ only depends on the endpoints $\phi(a)$ and $\phi(b)$ where $\phi:[a,b]\to U$. I have considered the hint, but don't know what to do with it. It sounds almost definitional to me.
Let me first elaborate on the hint. Let's take a one-form $\omega = g(x) \, dx$ on the real line $\mathbb{R}$. Let $x_0$ be any point in $\mathbb{R}$. Define a function $f(x)$ by $$f(x) = \int_{x_0}^x \omega = \int_{x_0}^x g(x) \, dx.$$ Here's a question for you: by the second fundamental theorem of calculus, what is $\frac{df}{dx}$? Using that, what is $df$?
Now let's go to $U \subset \mathbb{R}^n$. Again, let $x_0$ be any point in $U$. Given $\omega$, let's try to define $f(x)$ in precisely the same way as above: $$f(x) = \int_{x_0}^x \omega.$$ Here are a couple of things for you to think about:
Edit: Here's what you should do: let $f(x) = \displaystyle \int_{\phi_x} \omega$, where $\phi_x: [a,b] \to U$ is any path such that $\phi_x(a) = x_0$ and $\phi_x(b) = x$. Such a path exists because $U$ is connected. (We probably want $\phi_x$ to be piecewise smooth, but let's brush that under the rug.) The point is that it doesn't matter which path $\phi_x$ we choose, because of the property of $\omega$ that we were given. (Note that a different choice of the starting point $x_0$ would only change $f$ by a constant and thus not change $df$; do you see why?)