Find $f(x,y )$ such that $f_{x},f_{y},f_{yx}$ are continuous,but $f_{xy}$ is not

In Dieudonné's Elements of Analysis, volume 1, you'll find an exercise which states that if $f$ is $C^1$ and $f_{xy}$ exists and is continuous on an open set $U$, then $f_{yx}$ exists and equals $f_{xy}$ on $U$. (So, of course, $f_{xy}$ is continuous as well.)

Here's a sketch of the proof. As in the usual proof that a $C^2$ function has equal mixed partial derivatives, we define the second difference function $$\Delta(h,k) = f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b).$$ By applying the usual mean value theorem to $$g(t)=f(t,b+k)-f(t,b),$$ we see that $\Delta(h,k)=\big(f_x(\xi,b+k)-f_x(\xi,b)\big)h$ for some $\xi$ between $a$ and $a+h$. It follows that $\Delta(h,k)=f_{xy}(\xi,\eta)hk$ for some $\eta$ between $b$ and $b+k$. Because $f_{xy}$ is continuous at $(a,b)$, we infer that, for any $\epsilon>0$, there is $\delta>0$ so that whenever $|h|,|k|<\delta$ we have $$\big|\Delta(h,k)-f_{xy}(a,b)hk\big|<\epsilon|hk|.$$

Now we consider $q(t)=f(a+h,t)-f(a,t)$ and rewrite $\Delta(h,k) = q(b+k)-q(b) = q'(\tau)k$ for some $\tau$ between $b$ and $b+k$. This tells us that $\Delta(h,k)=\big(f_y(a+h,\tau)-f_y(a,\tau)\big)k$, and so, for $0<|h|,|k|<\delta$ we have $$\left|\frac{\Delta(h,k)}{hk}-f_{xy}(a,b)\right| = \left|\frac{f_y(a+h,\tau)-f_y(a,\tau)}h-f_{xy}(a,b)\right|<\epsilon.$$ Letting $k\to 0$ (so $\tau\to b$) and using continuity of the first partial derivatives, we obtain $$\left|\frac{f_y(a+h,b)-f_y(a,b)}h-f_{xy}(a,b)\right|\le\epsilon,$$ from which we conclude that $f_{yx}(a,b)=f_{xy}(a,b)$, as desired.

Note that to prove equality of the mixed partials at $(a,b)$ uses only continuity of $f_{xy}$ at $(a,b)$. :)

Sadly, however interesting, this leaves open the original question about whether $f_{yx}$ is continuous at $(a,b)$ knowing just that $f_{xy}$ is continuous at $(a,b)$.

EDIT: It follows from the Baire Category Theorem that if $F$ is differentiable, the set of discontinuities of $F'$ is of first category; in particular, it cannot be a dense open set. This still does not quite answer the question, I guess. It is not necessary that $f_{xy}$ be continuous in order for $f_{yx}$ to equal $f_{xy}$ (or to be defined for other reasons), but it sure seems unlikely that there is any counterexample. Anyone else?