Comment by WHLin on Is it true that if $(f\circ p)'(0) = \nabla f(0)p'(0)$ for all smooth paths $p$, then $f$ is differentiable at $0$?

I think it is correct. My intuition is: maybe you could consider radial coordinate, and use the path argument to show that F(p(t)) -> 0 as t ->0 "uniformly" within the neighborhood of 0. Then, the limit of F(h) under radial coordinate reduced to 1D problem (as long as the convergent speed is uniform).