One-sided derivative of composition function $f : V \subset\mathbb R^n \to \mathbb R$ is differentiable, $g : [0,1] \to  V$ a continuous function. Given $g(1)=p, Df(p)=0$, and that $f\circ g $ is left differentiable, can we deduce that the left derivative of $f \circ g $ at $1$ is $0$? If not, is there any counterexample? 

That is in general wrong.
Let $f(x) = x^2$ and $g(x) = \sqrt{1 - x}$ for $x\in[0, 1]$. 
Then, we have
$$ (f\circ g)'(x) = \frac{d}{dx}( 1 - x ) = -1. $$
