How can I prove $∇U(x).D_m x(p,m)= \text{shadow price}$?
Assuming certain regularity conditions, the first order conditions for 
$$
\max_{x, \lambda} U(x) - \lambda (p \cdot x - m)
$$
are
\begin{align*}
&amp;D_{x}U(x(p, m)) - \lambda p = 0 
\\
\text{and} \quad
&amp; p \cdot x(p, m) - m = 0.
\end{align*}
Moreover $x(p, m)$ will be differentiable with respect to $m$ at $(p, m)$, and this fact together with the second equation implies
$$
p \cdot D_{m}x(p, m) - 1 = 0.
$$
So, by the first equation,
$$
D_{x}U(x(p, m)) \cdot D_{m}x(p, m)
=
\lambda p \cdot D_{m}x(p, m)
=
\lambda.
$$
The derivative $D_{x}U(x(p, m))$ of the utility function at $x(p, m)$ corresponds to $\nabla U(x)$ in your notation.