Rudin states that $1$-form $xdy$ is not the derivative of any $0$-form.
By contradiction, suppose that that exists $0$-form whose derivative is $xdy$. Then $f\in C'$ and $df=xdy$. But $d^2f=dx\land dy\neq 0$. But how to apply part (b) of theorem 10.20? For right application our $f$ must be from $C''$ but it's from $C'$.
I would be very thankful for any answer.
Actually, if $f\in C'$ and satisfies $$df=xdy$$ it means that $$\dfrac{\partial f}{\partial x}(x,y)=0 \text{ and } \dfrac{\partial f}{\partial y}(x,y)=x.$$ And as you can see each partial derivative is again of class $C'$ and thus $f$ has to be of class $C''$.
Now, stating that $d^2f\neq 0$ is just an other way to notice that the Schwarz theorem doesn't apply : $$\dfrac{\partial^2 f}{\partial x\partial y}\neq \dfrac{\partial^2 f}{\partial y\partial x}$$ which is a contradiction.