On equivalent ways to obtain the modulus of the gradient. Recently I was following a seminar of one of the great living Italian mathematicians: Luigi Ambrosio (recently his student Alessio Figalli won the fields medal). One of his slides was dealing with the Eulerian/Lagrangian duality (a concept I was not so familiar with):

I have two questions: Could one provide some intuition as to why these 3 ways of computing the modulus of the gradient are the same and could I have a reference to a proof of this fact? I am interested in all three equivalencies.
 A: You didn't include any hypotheses, but of course we need to assume $f$ is differentiable at $x$ (or else you can have, for example, a function with $\nabla f(x)=0$ but some $\partial_v f(x)\ne 0$). But then it's standard from multivariable calculus that
$$\partial_v f(x) = \nabla f(x)\cdot v,$$
and so you achieve the maximum directional derivative by taking the unit vector $v$ in the direction of $\nabla f(x)$. This is the second formula (the sup is actually a max).
The third formula arises by computing directional derivatives as derivatives along curves. Indeed, if $\gamma$ is any curve with $\gamma(0)=x$ and $\gamma'(0)=v$, then by the chain rule we have 
$$\partial v f(x) = (f\circ \gamma)'(0) = (d_xf)(\gamma'(0)).$$
You can relate the first formula to the second, for example, by passing to a sequence $y_k\to x$ with $\dfrac{y_k-x}{\|y_k-x\|}\to v$, and then the quotient in question is approaching the directional derivative in direction $v$. Thus, you get the limsup by taking such a sequence that gives the optimal $v$.
EDIT: Note that the definition of the derivative $d_x f$ tells us that
$$f(y_k)-f(x) = d_xf(y_k-x) + o(\|y_k-x\|),$$
and so
$$\frac{f(y_k)-f(x)}{\|y_k-x\|} = d_x f\big(\frac{y_k-x}{\|y_k-x\|}\big) + \epsilon,$$
where $\epsilon\to 0$ as $y_k\to x$.
This means that when we chose the sequence $y_k\to x$ so that $\dfrac{y_k-x}{\|y_k-x\|}\to v$, we see that
$$\frac{f(y_k)-f(x)}{\|y_k-x\|)} \to d_xf(v) = \partial_v f(x),$$
as desired.
