Do these two properties guarantee that $f:A\to\mathbb{R}^n$ is differentiable at $x\in A^{\circ}\subset\mathbb{R}^m$? Consider the following two properties for a function $f: A \to \mathbb{R}^n$ where $A\subseteq \mathbb{R}^m$ at a point $x\in A^{\circ}$:
(a) For every $h\in\mathbb{R}^m$ $$\lim_{\lambda\to0}\dfrac{1}{\lambda}\left(f(x+\lambda h)-f(x)\right)$$
exists, where $\lambda\in\mathbb{R}$.
(b) If we set $$F(h)=\lim_{\lambda\to0}\dfrac{1}{\lambda}\left(f(x+\lambda h) - f(x)\right),$$
then $F:\mathbb{R}^m\to\mathbb{R}^n$ is a linear transformation.
My question is whether (a) and (b) together imply that $f$ is differentiable at $x$? I think the answer is no but am having trouble formulating a counter-example.
 A: The answer is no.
Probably the simplest counterexample is
$$f(x)=\begin{cases} x & y=x^2 \\ 0 & y\neq x^2\end{cases}$$
at the origin. Note that $f(0)=0$.
Property (a), which says that all directional derivatives exist at $x$, is satisfied, as you can easily check; with $h=(h_1, h_2)$, we have
$$\lim_{\lambda\to 0}\frac{1}{\lambda}f(\lambda h_1, \lambda h_2)=0$$
because as we shrink $\lambda$, eventually the scaled vector $\lambda h$ does not lie on the parabola $y=x^2$.
Property (b) is also satisfied, trivially: $F(h)\equiv 0$, which is a linear function of $h$.
But $f$ is not differentiable at the origin because
$$\lim_{h\to0}\frac{f(h)-F(h)}{\|h\|}=\lim_{h\to0}\frac{f(h)}{\sqrt{h_1^2+h_2^2}}$$
does not exist. If we approach the origin nonlinearly along the curve $y=x^2$, the difference quotient tends to $\pm1$ (depending on the sign of $h_1$); otherwise it tends to $0$.
Here I'm using the fact that if the derivative exists, it must be the function $F$.

How did I think up this example? I created a simple non-differentiable kink along a nonlinear curve. Differentiability in several variables is subtle because limits in serval variables are subtle: the limit must exist not only along lines but no matter how we approach the point.
