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Set $r=\sqrt{x^2+y^2}$. The bound $|xy|\le r^2/2$ and $|x^2-y^2|\le r^2$ give $|f(x,y)|\le r^2/2$. Hence $f\to 0$ at the origin, proving continuity, and $|f(x,y)|/r\to 0$, proving Fréchet differentiability there with derivative equal to the zero linear map.

Along the coordinate axes, direct difference quotients giveThese formulas include the origin. Therefore the two mixed partial derivatives differ:The explicit operator order removes any ambiguity about the notation for mixed derivatives. Also $f_{xx}(0,0)=f_{yy}(0,0)=0$, since the first derivatives restricted to their own axes vanish identically.

All four second-order partial derivatives are discontinuous at the origin. Away from it,Along $x=y\ne 0$ these are one and minus one, rather than their zero values at the origin. Meanwhile ${\partial }_y{\partial }_{x}f(x,0)=1$ for $x\ne 0$, whereas its value at the origin is minus one; and ${\partial }_x{\partial }_{y}f(0,y)=-1$ for $y\ne 0$, whereas its origin value is one. This gives a concrete unequal mixed partial derivatives example despite differentiability of the original function.

False. Unequal mixed partial derivatives can exist when the continuity hypotheses of Clairaut's theorem fail.