Is there a 0-form $\tau$ with $d\tau=\omega$? 

Consider the 1-form 
    $$
\omega=(x^2-yz)dx+(y^2-xz)dy-xydz.
$$
    Does a 0-Form $\tau$ on $\mathbb{R}^3$ exist which fullfils $d\tau=\omega$?


Hello, my simple answer is: YES, because on $\mathbb{R}^3$ there exists an antiderivative for $\omega$, because my calculations showed that $\omega$ is closed and furthermore $\mathbb{R}^3$ is path connected.
This antiderivative is a suitable 0-Form here.
Am I right with this argumentation?
 A: A cohomological argument: since $H^1(\mathbb{R}^3)=0$, any degree $1$ cocycle must be a degree $1$ coboundary.  If you've shown that $\omega$ is closed, it must be the coboundary of some $0$ form $\tau$.
Or, as suggested in the comments, you can just compute.
A: A standard multivariate calculus argument is as follows. 
For $0$ form $\tau = \phi(x,y,z)$ so that $d\tau = \omega$, we just wanna find a $\phi$ satisfying:
$$
\frac{\partial \phi}{\partial x} = x^2−yz,
\\
\frac{\partial \phi}{\partial y} = y^2−xz,
\\
\frac{\partial \phi}{\partial z} =−xy.
$$
Integrating the first with respect to $x$ gives :
$$
\phi = \frac{x^3}{3} - xyz + g(y,z).
$$
Taking partial $y$:
$$
\frac{\partial \phi}{\partial y} = −xz +\frac{\partial g}{\partial y} = -xz + y^2.
$$
Integrating $g$ with respect to $y$: 
$$
g = \frac{y^3}{3} + h(z) \implies \phi = \frac{x^3}{3} - xyz + \frac{y^3}{3} + h(z).
$$
Then taking derivative with respect to $z$ and find $h(z)$.

To translate Jared's answer, simply checking $\nabla \times (x^2−yz,y^2−xz,-xy) = 0$ in all $\mathbb{R}^3$ suffices to guarantee the existence of $\tau$.
