Different notions of differentiability The following is somewhat unclear to me. Let $X$, $Y$ be locally convex vector spaces, let $f: X \supseteq U \longrightarrow Y$ be a (nonlinear) continuous map. Then one can say that $f$ is $C^1$ if 
a) the difference quotient 
$$ Df(x)\xi := \lim_{\epsilon \longrightarrow 0} \frac{f(x + \epsilon \xi) - f(x)}{\epsilon}$$
exists for all $x \in U$, $\xi \in X$ and if
$$Df: U \times X \longrightarrow Y$$
is continuous; or
b) $f$ maps $C^1$ curves in $X$ to $C^1$ curves in $Y$.
Question: Are the notions a), b) equivalent?
As far is understood, they are not in general, right? But for which spaces are they equivalent?
Addition: If $X$ and $Y$ are not Banach spaces, it seems too strong a notion to require the map $Df$ to be continuous from $X$ to $L(X, Y)$. Why is this so?
 A: b) does not imply a) even when $X=Y=\mathbb R^2$. It's easier to describe the counterexample in complex notation:
$$f(r\exp(i\theta))  = r\exp(i\theta+i\sin 2\theta),\quad r,\theta\in\mathbb R  \tag1$$
It is important to notice that (1) applies both when $r<0$ and when $r>0$, giving consistent result because $\sin 2\theta$ is $\pi$-periodic.
The map (1) is not differentiable at the origin, because it is degree 1 homogeneous but not linear. 
On the other hand, it maps $C^1$ curves to $C^1$ curves. Indeed, let $\Gamma:(-1,1)\to \mathbb R^2\approx \mathbb C$ be a $C^1$-curve. For definiteness assume $\Gamma(0)=0$. Let $\rho e^{i\phi}=\Gamma'(0)$; here $\rho>0$. Write $\Gamma(t)=t \rho e^{i\phi}+o(t)$ and plug this into $f$. Since $f$ is Lipschitz, we gave $$f\circ \Gamma(t)=f(t \rho e^{i\phi})+o(t) = t\rho e^{i\phi+i\sin 2\phi}+o(t) \tag2 $$  Hence, $f\circ \Gamma$ has a tangent at $t=0$.  Its tangent at the points where $\Gamma\ne 0$ has argument $\arg \Gamma'(t)+2\sin \arg \Gamma(t)$ which is consistent with (2): that is, the tangent direction is continuous along the curve.
