= Solution
Set $r=\sqrt{x^2+y^2}$. The bound $|xy|\leq r^2/2$ and $|x^2-y^2|\leq r^2$ give $|f(x,y)|\leq r^2/2$. Hence $f\to0$ at the origin, proving <continuity>, and $|f(x,y)|/r\to0$, proving <Fréchet differentiability> there with \b[derivative equal to the zero linear map].
Along the coordinate axes, direct difference quotients give
$$
\boxed{f_x(0,y)=-y,\qquad f_y(x,0)=x.}
$$
These formulas include the origin. Therefore the two <mixed partial derivatives> differ:
$$
\boxed{\partial_y(\partial_x f)(0,0)=-1,\qquad\partial_x(\partial_y f)(0,0)=1.}
$$
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,
$$
f_{xx}(x,y)=\frac{4xy^3(-x^2+3y^2)}{(x^2+y^2)^3},\qquad f_{yy}(x,y)=-\frac{4yx^3(-y^2+3x^2)}{(x^2+y^2)^3}.
$$
Along $x=y\ne0$ these are one and minus one, rather than their zero values at the origin. Meanwhile $\partial_y\partial_xf(x,0)=1$ for $x\ne0$, whereas its value at the origin is minus one; and $\partial_x\partial_yf(0,y)=-1$ for $y\ne0$, whereas its origin value is one. This gives a concrete <unequal mixed partial derivatives> example despite differentiability of the original function.
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