Find the gradient of $f^*(x)=\langle (\nabla f)^{-1}(x),x)\rangle-f( (\nabla f)^{-1}(x))$ for $x \in \mathbb{R^n}$ I am stuck at the following exercise which serves as a preparation for the upcoming exam:

Let $U \subset \mathbb{R}^n$ be open and $f \in C^2(U, \mathbb{R})$ such that $\det Hf(x) \neq 0, \forall x \in U$ and $\nabla f: U \to \nabla f(U) =:V$ bijective.

Under the above assumptions the function $f^*: V \subset \mathbb{R}^n \to \mathbb{R}$ given by $$f^*(u):=\langle (\nabla f)^{-1}(u),u)\rangle-f( (\nabla f)^{-1}(u)) \tag{*}$$
is clearly well defined and $f^* \in C^1(V, \mathbb{R})$ thanks to the Hessian Matrix of $f$ being invertible and thus $\nabla f$ is a local diffeomorphism. The bijective nature of $\nabla f$ makes $(\nabla f)^{-1}$ a class $C^1$ function.

My problem: I am supposed to find the gradient of $f^*$, in fact I am supposed to verify the identity: $$(\nabla f^*)(u)=(\nabla f)^{-1} (u) \tag{$\alpha$} $$
My approach: Let $u=(u_1, \dots , u_n) \in \mathbb{R}^n$ then it makes sense to only see if the statement in ($\alpha)$ holds for an arbitrary $i \in \lbrace 1, \dots , n \rbrace$. 
My problem is that I know nothing about how $(\nabla f)^{-1}$ looks like, so I thought the best attempt would be to just let it be in an abstract form.
Let $h(u):=\langle (\nabla f)^{-1}(u),u)\rangle$ then I assume $h(u)$ looks like $$ h(u)=(\nabla f)_1^{-1}(u)*u_1 + \dots + (\nabla f)_i^{-1}(u)*u_i + \dots + (\nabla f)_n^{-1}(u)* u_n \\ \implies \frac{\partial h(u)}{u_i}=\frac{\partial(\nabla f)_1^{-1}(u)}{\partial u_i}u_1+ \dots + \frac{\partial(\nabla f)_i^{-1}(u)}{\partial u_i}u_i + (\nabla f)_i^{-1}(u)+ \dots + \frac{\partial (\nabla f)_n^{-1}(u)}{\partial u_i}u_n$$
If I have made no mistakes, all there is left to do is to check the second expression in (*) and hope that it annihilates the partially differentiated terms in my expression above. But all I get is: $$\frac{\partial f(( \nabla f^{-1})(u))}{\partial u_i}= \frac{\partial f}{\partial u_i}( \nabla f)^{-1}(u) \cdot \frac{\partial(\nabla f)_i^{-1}}{\partial u_i}(u)$$
So it would be nice if the first term would simplify to $u_i$ but I really don't see it.
Any help or simplification would be appreciated.
 A: I propose a solution which is coordinates free (the proof will hold in infinite dimension).
By definition of the differential, if $f:\mathcal{U}\longrightarrow\mathbb{R}$ is differentiable at the point $u\in\mathcal{U}$, then for all $h\in\mathbb{R}^n$ such that $u+h\in\mathcal{U}$, we have
$$f\left(u+h\right)
=f\left(u\right)+\underbrace{\mathrm{d}f\left(u\right)}_{\in\mathcal{L}\left(\mathcal{U};\mathbb{R}\right)}\left[h\right]+o\left(h\right)$$
where $o\left(h\right)=\|h\|\varepsilon\left(h\right)$ with $\varepsilon\left(h\right)\rightarrow 0$ when $h\rightarrow0$.
Now we can define the gradient $\nabla f\left(u\right)\in\mathbb{R}^n$ of $f$ at the point $u\in\mathcal{U}$ as
$$\left\langle\nabla f\left(u\right),h\right\rangle
=\mathrm{d}f\left(u\right)\left[h\right]\in\mathbb{R}.$$
This definition holds for any Hilbert space.
Now we differentiate $f^*$ assuming that $\left(\nabla f\right)^{-1}$ is differentiable on $\mathcal{U}$ :
$$f^*\left(u+h\right)
=\left\langle\left(\nabla f\right)^{-1}\left(u+h\right),u+h\right\rangle-f\left(\left(\nabla f\right)^{-1}\left(u+h\right)\right)$$
$$=\left\langle\left(\nabla f\right)^{-1}\left(u\right)+\mathrm{d}\left(\left(\nabla f\right)^{-1}\right)\left(u\right)\left[h\right]+o\left(h\right),u+h\right\rangle-f\left(\left(\nabla f\right)^{-1}\left(u\right)+\mathrm{d}\left(\left(\nabla f\right)^{-1}\right)\left(u\right)\left[h\right]+o\left(h\right)\right)$$
$$=\left\langle\left(\nabla f\right)^{-1}\left(u\right),u\right\rangle
+\left\langle\left(\nabla f\right)^{-1}\left(u\right),h\right\rangle
+\left\langle\mathrm{d}\left(\left(\nabla f\right)^{-1}\right)\left(u\right)\left[h\right],u\right\rangle
-f\left(\left(\nabla f\right)^{-1}\left(u\right)\right)
-\mathrm{d}f\left(\left(\nabla f\right)^{-1}\left(u\right)\right)\left[\mathrm{d}\left(\left(\nabla f\right)^{-1}\right)\left(u\right)\left[h\right]\right]
+o\left(h\right)$$
$$=f^*\left(u\right)
+\left\langle\left(\nabla f\right)^{-1}\left(u\right),h\right\rangle
+\left\langle\mathrm{d}\left(\left(\nabla f\right)^{-1}\right)\left(u\right)\left[h\right],u\right\rangle
-\mathrm{d}f\left(\left(\nabla f\right)^{-1}\left(u\right)\right)\left[\mathrm{d}\left(\left(\nabla f\right)^{-1}\right)\left(u\right)\left[h\right]\right]
+o\left(h\right).$$
Hence, we have :
$$\mathrm{d}\left(f^*\right)\left(u\right)\left[h\right]
=\left\langle\underbrace{\left(\nabla f\right)^{-1}\left(u\right)}_{\in\mathbb{R}^n},h\right\rangle
+\left\langle\underbrace{\mathrm{d}\left(\left(\nabla f\right)^{-1}\right)\left(u\right)\left[h\right]}_{\in\mathbb{R}^n},u\right\rangle
-\mathrm{d}f\underbrace{\left(\left(\nabla f\right)^{-1}\left(u\right)\right)}_{\in\mathbb{R}^n}\underbrace{\left[\mathrm{d}\left(\left(\nabla f\right)^{-1}\right)\left(u\right)\left[h\right]\right]}_{\in\mathbb{R}^n}\in\mathbb{R}.$$
By definition of the gradient in $\mathbb{R}^n$, we can write the last term above as
$$\mathrm{d}f\left(\left(\nabla f\right)^{-1}\left(u\right)\right)\left[\mathrm{d}\left(\left(\nabla f\right)^{-1}\right)\left(u\right)\left[h\right]\right]
=\left\langle \nabla f\left(\left(\nabla f\right)^{-1}\left(u\right)\right),\mathrm{d}\left(\left(\nabla f\right)^{-1}\right)\left(u\right)\left[h\right]\right\rangle$$
$$=\left\langle u,\mathrm{d}\left(\left(\nabla f\right)^{-1}\right)\left(u\right)\left[h\right]\right\rangle
=\left\langle \mathrm{d}\left(\left(\nabla f\right)^{-1}\right)\left(u\right)\left[h\right],u\right\rangle$$
therefore, we finally find
$$\left\langle\nabla \left(f^*\right)\left(u\right),h\right\rangle
=\mathrm{d}\left(f^*\right)\left(u\right)\left[h\right]
=\left\langle\left(\nabla f\right)^{-1}\left(u\right),h\right\rangle$$
and this is true for all $h\in\mathbb{R}^n$ such that $u+h\in\mathcal{U}$, for all $u\in\mathcal{U}$. Hence, we proved that
$$\nabla \left(f^*\right)=\left(\nabla f\right)^{-1}$$
as wanted.
