Transformation mapping vector to a gradient is diffeomorphism $f:U\to \mathbf{R}$ is $C^2$, $U$ is open subset of $\mathbf{R}^n$, for every $u\in U$ $D^2f(u)$ is positive definite. How do I show that the map $x\mapsto \nabla f(x)$ is a diffeomorphism?
 A: To nitpick your statement, this is not true if $U$ is disconnected. For instance, take $U = (0,1) \cup (2,3) \subset \mathbb{R}$ and the function $f:U \to \mathbb{R}$ defined by $f(x) = x^2$ if $x \in (0,1)$ and $f(x)=(x-2)^2$ if $x \in (2,3)$.
Obviously $f$ is $C^2$ on the domain $U$ and $D^2f(u) = f''(u) = 2$ on each of the separate parts of $U$. Thus, we can think of $D^2f(u)$ as the constant $1 \times 1$ positive-definite "matrix" $(2)$. So clearly, my function and its derivatives satisfy the requirements in your question.
However, $x \mapsto \nabla f(x)=Df(x)=f'(x)$ still fails to be a diffeomorphism for the simple reason that it is not injective. For example, you can calculate and show that $f'(0.5) = f'(2.5)$.
Nitpicking aside, this fact is true if $U$ is an open convex subset of $\mathbb{R}^n$ (which is connected):
For convenience, I use the notation $Df$ instead of $\nabla f$.
Well, you first have to show that $Df$ is an injection. You get surjection for free since I am assuming your codomain is the image set $Df(U)$.
To show the sought for injectivity, suppose $Df(u) = Df(v)$ for some $u, v \in U$. We have to somehow prove $u = v$. To do this, remember that $U$ is convex in $\mathbb{R}^n$. So, by definition, the straight line path $$\lambda(t) = u+t\Delta u \qquad \Delta u = v - u \quad t \in [0,1]$$ from $u$ to $v$, is entirely contained within $U$. So it makes sense to talk about $Df(\lambda(t))$ and $D^2f(\lambda(t))$.
Now, define the real-valued function $I(t) = \langle Df\lambda(t) - Df(u), \Delta u \rangle$ and note that its derivative is $$I'(t)=\langle D^2f\lambda(t) \cdot D\lambda(t), \Delta u \rangle = \Delta u^T \cdot D^2f\lambda(t) \cdot \Delta u$$ Since we assumed $Df(u)=Df(v)$, we get $I(0)=I(1)=0$. So, the mean value theorem gives us a $c \in (0,1)$ such that $I'(c)= \Delta u^T \cdot D^2f\lambda(c) \cdot \Delta u = 0$. As $D^2f$ is positive-definite, this is only possible if $\Delta u = v-u=0$.
Thus, $Df:U \to Df(U)$ is a differentiable bijection. To show that its inverse is also differentiable, just note that $D^2f(u)$ is invertible by positive-definiteness and apply the inverse function theorem.
