Is $\omega = dU = sin(x+y)dx+cos(x+y)dy$ an exact form?

Question

In my thermodynamics homework I should prove that $dU = sin(x+y)dx+cos(x+y)dy$ is a function of state. Which means it's integration over any path be constant or in other word $dU$ should be an exact form. I used the Poincare Lemma and had the following calculus:
$$d\omega =(D_1sin(x+y)dx+D_2sin(x+y)dy)\wedge dx+(D_1cos(x+y)dx+D_2cos(x+y)dy)\wedge dy$$ $$= -(cos(x+y)+sin(x+y))dx\wedge dy$$ which is not zero. So $\omega$ isn't a closed so an exact so a function of state. Can you help me please?

Answer 1

$\large\mbox{It's not !!!}$

$$ {\partial U \over \partial x} = \sin\left(x + y\right)\,, \quad U = -\cos\left(x + y\right) + \phi\left(y\right) $$

$$ {\partial U \over \partial y} = \sin\left(x + y\right) + \phi'\left(y\right) \color{#ff0000}{\LARGE\not=} \cos\left(x + y\right) $$