A simple question about the difference between directional derivatives and derivatives As the topic, let $\mathbf{u}=u\hat{i}+v\hat{j}$. 
Why $$D_ug(0,0) = \lim_{h\rightarrow0}\frac{g(hu,hv)-g(0,0)}{h}$$ while
$$Dg(0,0)= \lim_{(h,k)\rightarrow (0,0)}\frac{g(h,k)-L(0,0)}{\sqrt{h^2+k^2}},$$we don't need to use the linearization in the case of directional derivatives.
 A: I will assume that $L(0,0) = g(0,0)$; otherwise, the error in the second definition will be elsewhere, but I will need the definition of $L$ to pinpoint it.
The problem is that the second definition is false (in that it does not match with the usual definition of the derivative). For instance, if we take $g(h,k) = h$, we get:
$$ Dg (0,0) = \lim_{(h,k) \to (0,0)} \frac{h}{\sqrt{h^2+k^2}},$$
but this limit does not exist: if $h=0$ and $k$ goes to $0$, then the limit is $0$, but if $k=0$ and $h$ goes to $0$ then this limit is $\pm 1$. So, with this definition, even this function $g$ would not be differentiable!
The definition of the derivative $Dg (0,0)$ of $g$ in $(0,0)$ is that it is (if it exists) a linear operator from $\mathbb{R}^2$ to $\mathbb{R}$, such that for all vectors $\mathbf{u}$:
$$ \lim_{(h,k) \to (0,0))} \frac{g(h,k) - g(0,0) - Dg (0,0) \mathbf{u}}{\sqrt{h^2+k^2}} = 0,$$
or equivalently:
$$g(h,k) =_{(0,0)} g(0,0) + Dg (0,0) (h,k) + o (\|(h,k)\|).$$
If the derivative of $g$ exists with this definition, then $D_{\mathbf{u}} g (0,0) = Dg (0,0) \mathbf{u}$ [exercise], so that this definition of the derivative and the definition of the directionnal derivative are coherent. Note, however, that the converse is not true: the fact the directionnal derivatives exist do not imply that a ("full") derivative exists.
