Using Clairaut's Theorem to show $\beta=yzdx -xzdy +xydz$ is not a differential of a function? $\beta=yzdx -xzdy +xydz$ 
is the one form and I'm struggling to show how this is not a differential of a function using Clairaut's Theorem.
Suppose $f$ is the said function then it will intake the $3$ variables $(x,y,z)$, is it true to say that if all second order partial derivatives only with respect to $x$ and $y$ (or equivalently any pair) are continuous then they are all equal? 
Or because $f$ is a function of $3$ variables then can we only say that if all the third order partial derivatives (including all variables) are continuous then they are equal? 
Going to the example with the one form $\beta$, how to show $\beta$ is not the differential of a function?
My attempt:
If $\beta$ is the differential of a function then:
$f_x = yz$ ;    $f_y = -xz$ ; $f_z = xy$ 
Since $f_{xy} \ne f_{yx}$ this is sufficient to show $\beta$ is not a differential of a function(I know this would suffice for a function of $x$ and $y$ only)? 
Or must we show that all third order partial derivatives (including all variables) are continuous and that they are not all equal?
 A: Doing this for only two variables is enough. Recall that a form is called exact if it is the differential of another form, and closed if its differential is $0$. Because the differential operation squares to $0$, every exact form is closed. Thus, one way to show that a form is not the differential of another form is to show that its differential is nonzero.
Let's apply the above rule to the form you're working with. To see that $\beta$ is not exact, it suffices to show that $\beta$ is not closed; i.e., it suffices to show that $d\beta \neq 0.$ It is easy to compute that
$$d\beta = -2z (dx \wedge dy) + 2x (dy \wedge dz),$$
and the above two-form is evidently nonzero. We conclude that $\beta$ is not the differential of a form.
Now, notice that the fundamental two-forms $dx \wedge dy$, $dy \wedge dz$, and $dx \wedge dz$ form a basis for the space of two-forms, so to show that $d\beta$ is not zero, it suffices to show that the coefficient of $dx \wedge dy$ in the coordinate expansion of $d\beta$ is nonzero. But if $\beta = df$, then the coefficient of $dx \wedge dy$ in $f_{xy} - f_{yx}$, which is zero if and only if the hypothesis of Clairaut's theorem is satisfied by the functions $g^z(x,y) = f(x,y,z)$ for every fixed value of $z$. So, yes, there is a precise sense in which it suffices only to consider two of the variables.
Note: The hypothesis of Clairaut's theorem is satisfied for the variables $x$ and $z$, but that does not mean that the form $\beta$ is the differential of a function. All we can say is that a form is not a differential of a function if the hypothesis of Clairaut's theorem is not satisfied for some pair of the variables.
