A diffeomorphism whose tangent map preserves dot products is an isometry. I'm having trouble solving the following problem.
If $F:\mathbb{R^3} \to \mathbb{R^3}$ is a diffeomorphism such that $F\ast$(the tangent map of $F$) preserves dot products, show that $F$ is an isometry. (Hint: Show that $F$ preserves lengths of curve segments and deduce that $F^{-1}$ does also.)
I'm trying to proceed as in the hint, but I'm stuck on the first step, showing the $F$ preserves lengths of curve segments. Moreover, how does showing that both $F$ and its inverse preserves length of curve segments show that it is an isometry. And how do I characterize the curve segments. I would appreciate any solutions, hints or suggestions.

$dF(u)\cdot dF(v) = u\cdot v \implies |dF(u)|^2 = |u|^2.$ So $dF$ is an isometry at each point. Suppose $\gamma : [0,1] \to \mathbb {R}^3$ is a nice curve from $a$ to $b.$ Then $F\circ \gamma$ is a nice curve from $F(a)$ to $F(b).$ The length of $F\circ \gamma$ is $\int_0^1|(F\circ \gamma)'(t)|\, dt = \int_0^1|dF(\gamma (t))[\gamma '(t)]|\, dt = \int_0^1|\gamma '(t)|\, dt.$ This last integral is the length of $\gamma.$ This and similar machinations should help finish the job. 
