Schwarz's theorem and a counterexample

Schwarz's theorem. If $f$ has continuous second partial derivatives in a neighbourhood of a point $(a,b)$, then $f_{xy}(a,b)=f_{yx}(a,b)$.

First partials at the origin. For $x\neq0$,
$$\frac{f(x,0)-f(0,0)}{x}=\frac{0}{x}=0,$$
so $f_x(0,0)=0$. Similarly $f_y(0,0)=0$.

Mixed partials at the origin.
$$f_{xy}(0,0)=\lim_{y\to0}\frac{f_x(0,y)-f_x(0,0)}{y},\qquad f_{yx}(0,0)=\lim_{x\to0}\frac{f_y(x,0)-f_y(0,0)}{x}.$$
A direct computation gives $f_{xy}(0,0)=-1$ and $f_{yx}(0,0)=1$.

Why no contradiction. The second partial derivatives are not continuous at $(0,0)$: along paths approaching the origin, $f_{xy}$ and $f_{yx}$ do not admit a limit equal to $-1$ and $1$ respectively. Schwarz's hypothesis ($C^2$ regularity near the point) fails, so the conclusion need not hold.

Conclusion. Schwarz's theorem is compatible with the existence of such counterexamples: only $C^2$ functions are guaranteed to have symmetric mixed partials.