is function differentiable iff directional derivative is linear Original definition

A function $f: A \to \mathbb{R}^n$, $A \subseteq \mathbb{R}^m$ is differentiable at a point $\mathbf a \in \mathbb R^m,$
  if there is a linear transformation $T$ such that
  $$
\lim_{\lVert \mathbf h\rVert \to 0}
\frac{f(\mathbf a+\mathbf h)-f(\mathbf a)-T_a(\mathbf h)}{\lVert \mathbf h\rVert} = \mathbf 0.
$$

My definition

A function $f: A \to \mathbb{R}^n$, $A \subseteq \mathbb{R}^m$ is differentiable at a point $\mathbf a \in \mathbb R^m,$
  if there is a linear transformation $T$ such that for any unit vector $\hat u$
$$
\lim_{t \to 0}
\frac{f(\mathbf a+t \hat u)-f(\mathbf a)}{t} = T_a(\hat u).
$$

Question: is there any difference between original definition and my definition
In the original definition, $\mathbf h$ can approach to $\mathbf 0$ by any trajectory; In my definition it can only be approached from certain direction, so my definition is weaker than the original definition. Substitute $\mathbf h = t \hat u$ to the original definition will get my definition.
If my definition is not true, can somebody provide a counter example ?
One may already noticed that
$$
\partial_{\hat u} f (a) = T_a(\hat u)
$$
Since $T$ is linear, we can assume (here $\nabla f$ is just a function, it happens to be equal to the gradient if exist):
$$
T_a(\hat u) = \hat u \cdot \nabla f(a)
$$
So my definition can be written as:
$$
\partial_{\hat u} f (a) = \hat u \cdot \nabla f(a) \Leftrightarrow f\text{ is differentiable at }a
$$
 A: For a counterexample try 
$$
f:\mathbb{R}^2 \longrightarrow \mathbb{R}
$$
with
$$
f(x,y) = \left\{
   \begin{array}
     11 &\text{if  } \:\:x=y^2\:\: \text{ and  }\:\:(x,y)\not=(0,0)\\
     0 & \text{otherwise}
   \end{array}
   \right.
$$
I don't want to prove everything for you. Try to imagine how this function looks like, why it's not differentiable in 0 (by general definition) and why partial derivatives exist on all directions (following your definition).
EDIT:
Your intuition is very good. The difference between the original definition of differentiability, and the one you proposed is in the fact that the original definition allows $h$ to approach the point $a$ in any trajectory, while you allow it only on linear trajectories.
This counterexample is a function that is 0 everywhere, except on a small set $A=\{(x,y)\:|\:x=y^2\} \setminus\{(0,0)\}$ where it is equal to 1. Looking at the function on this curve, it is obviously not continuous in 0. But it has directional derivatives in all directions in 0, since for every direction $\hat u$, the point $t\hat u$ is inside $\mathbb{R}^2\setminus A$ for $t$ small enough.
