Q: Find a smooth path along which a given function on the plane is not differentiable at the origin From Bamberg & Sternberg's A Course In Mathematics For Students of Physics, Exercise 6.1d:

Let $F(x,y) = \frac{x^3y}{x^2+y^2}$ for $(x,y) \neq (0,0)$ and
  $F(0,0)=0$. Invent a smooth curve $x=X(t)$, $y=Y(t)$ through the
  origin with $X(0)=Y(0)=0$, such that the function $G(t)=F(X(t),Y(t))$
  is not differentiable at the origin.

I don't think this is possible. $F$ is differentiable at the origin with $\mathrm dF_{(0,0)}=\mathbf 0$, I believe. (Converting to polar coordinates gives $F$ as $r^2$ times a bounded function, hence $o(r)$). The Chain Rule tells me that $F\circ \gamma$ for any smooth (i.e., differentiable) $\gamma(t)$ is also differentiable, so there can't be a smooth curve that satisfies the problem's conditions. What am I missing, if anything?

A: Looks like this exercise in the textbook has a typo somewhere, as I suspected.
