We were provided the following example in the context of Schwarz's theorem:
Let be $f:\mathbb{R}^2 \to \mathbb{R}$, given by: $$ f{x \choose y}=\left\{\begin{array}{ll} \frac {xy^3}{x^2+y^2}, & {x \choose y}\neq {0 \choose 0} \\ 0, & {x \choose y}={0 \choose 0}\end{array}\right. . $$
If we calculate the second order partial derivatives for each ${x \choose y} \neq {0 \choose 0}$ we get: $$ D_{12}f{x \choose y} = D_{21}f{x \choose y}. $$
However, at point ${0 \choose 0}$ we have: $$ D_{12}f{0 \choose 0} =1 \neq 0 = D_{21}f{0 \choose 0}. $$
Our lecture notes say that we now can deduce only from Schwarz's theorem, that both second order derivatives $D_{12}f{x \choose y}$ and $D_{21}f{x \choose y}$ cannot be continuous at point ${0 \choose 0} $.
I don't understand how we can deduce this fact only from Schwarz's theorem?
Couldn't it be possible that only one of the two second order derivatives is continuous?
