Prove that $d(df)=0$ for any $0$-form

Question

As here shown the exterior derivative of the exterior derivative of a $0$-form of class $C^r$ for $r\ge 2$ is $0$ but as you can show the function $f:\Bbb R^2\rightarrow\Bbb R$ defined by the equation $$ f(x,y):=\begin{cases}0\,\,\,\text{if}\,\,\,(x,y)=(0,0)\\\frac{x^3y-xy^3}{x^2+y^2},\,\,\,\text{otherwise}\end{cases} $$ for any $x,y\in\Bbb R^2$ is of class $C^1$ and it is such that $f_{yx}(0,0)=-f_{xy}(0,0)=1$ so that $d(df)=(f_{xy}-f_{y,x})dx\wedge dy\neq0$ but unfortunately when my text (and others too) define the exterior derivative $d$ did not say that $d(df)=0$ only if $f\in C^r$ for $r\ge 2$ but when it prove this property it use the Schwartz theorem so that I am forced to say that this result holds only in the case where $r\ge 2$ because it could be to exist a scalar function $f$ of class $C^1$ whose derivative is differentiable and such that some mixed partial derivatives are note equal in any point. So I ask how the facts actually stand. So could someone help me, please?