Some clarification needed on the Relation between Total Derivative and Directional Derivative I will consider here functions of several variables only.
If both directional derivative $D_{v}f(x)$  at  $x$  along  $v$   and  total derivative $D f(x)$ at  $x$  exist  then
$$D_{v}f(x)=Df(x)(v).$$
Existence of total  derivative ensures  that  of  directional derivative  in every direction but not the other way round. 
There are functions who have, at some point of the domain, directional derivative in every direction but not differentiable at that point i.e. the total derivative at that point does exist.
Now, all of my knowledge is theoretical. I cannot see the picture clearly, i.e.  the picture of the two kinds of derivative existing together, or one existing and not the other - how do these work?
I mean some geometrical interpretation for say $2$ or $3$ dimensional space would help.
I am so confused with this thing, I am not even sure if I have managed to convey my problem properly.
Please help with some clarification.
Thanks.

We had a conversation before about both of us studying multivariable calculus before. It seems that we are in similar chapters. So I want to help you if I can. First of all, we have to clarify your first statement. You say that if both directional derivative $D_{v}f(x)$ at $x$ along $v$ and total derivative $Df(x)$ at $x$ exist then 
$$D_{v}f(x)=Df(x)(v)$$
This is not always the case. Sometimes both total derivative and directional derivative may exist, but
$$D_{v}f(x) \neq Df(x)(v)$$
For example, consider the function
$$f(x,y) = \begin{cases} \frac{x^2y}{x^2+y^2} \text{ if $(x,y) \neq (0,0)$} \\ 0 \text{$\qquad$ if $(x,y) = (0,0)$}\end{cases}$$
For this function, $D_vf(x)=1/2$ for $x=(0,0)$ and $v=\langle 1,1 \rangle$. However, $Df(x)v=0$. As you see, knowing only total derivative or directional derivative is not helpful. You should not expect a relationship between them unless required criterions are met. 
They are related ($D_{v}f(x)=Df(x)(v)$) if $f$ is differentiable at $x$. So you need a differentiability criterion. The following theorem helps: If, for a function $f:U \to \mathbb R^m$, all partial derivatives exist and they are continous on $U$, then $f$ is differentiable on $U$.  
