Geodesic flow generated by Riemannian distance function

This is an exercise in AC da Silva's Lectures onn Symplectic Geometry; I am having trouble showing the following.

$(X,g)$ is a geodesically complete manifold, and $f: X \times X \to \mathbb{R}$ is given by $f(x,y) = - \frac{1}{2} d(x,y)^2$ where $d$ is the Riemannian distance function (i.e. the infimum of arc-lengths of piecewise smooth curves joining $x$ to $y$).

I need to show that under the identification of $TX$ and $T^*X$ by the metric $g$, the symplectomorphism generated by $f$ coincides with the map $TX \to TX$ given by $(x,v) \to (\exp(x,v)(1), \frac{d}{dt} \exp(x,v) (1))$.

This amounts to solving the equations $g_x(v, \cdot ) = d_x f (\cdot )$ and $g_y (w, \cdot) = -d_y f(\cdot)$ for $(y, w)$ with fixed $(x,v)$, where $d_x f, d_y f$ are components of $df_{(x,y)}$ where $T^*_{(x,y)} (X \times X) \simeq T_x^* X \times T_y ^* X$.

I am having trouble in trying to solve the first equation, $g_x (v, \cdot) = d_x f(\cdot)$ for $y$, where both are members of $T_x^* X$. (The answer should be $y = \exp(x,v)=1$.) In particular, when $g$ (and hence $d$) is not concretely given, how should I go about finding $df$ at a fixed point $(x,y)$? Thank you for any help.

Given $$(x,v) \in TX$$, let $$\exp_x(v): \mathbb R \to X$$, $$t \mapsto \exp_x(tv)$$ be the unique geodesic with initial conditions

$$\begin{cases} \exp_x (0)= x \\ \frac{d}{dt}\big|_{t=0}\exp_x (tv) = v \end{cases}$$ Consider the map

\begin{aligned}\tilde g_x : T_xM &\to T_x^*M \\ v &\mapsto g_x (v, \cdot) \end{aligned} We need to solve

$$\begin{cases} \tilde g_x (v) = \xi_i = d_xf \\ \tilde g_x(w) = \eta_i = - d_yf \end{cases} \iff \begin{cases} g_x (v, \cdot) = d_xf (\cdot) &(\star) \\ g_x(w, \cdot) =- d_yf (\cdot) &(\star \star) \end{cases}$$

First we solve $$(\star)$$ for $$y \in X$$. Since $$X$$ is geodesically convex there exists $$u \in T_xM$$ such that $$\exp_x (u) = y$$.

Claim: For $$u,v \in T_x M$$ $$\frac{d}{dt}\bigg|_{t=0} \left(-\frac{1}{2} d(\exp_x(tv), \exp_x (u))^2\right) = g_x(u,v)$$

Proof: Let $$s \in \mathbb R$$ be sufficiently small so that $$\exp_x(su)$$ is contained in a geodesic ball centered at $$x$$ and $$\exp_x(u) = y \in \partial_sB(x)$$ (see Figure 1). Then we have that $$d(\exp_x(su) , \exp_x(u)) = |1-s||u|$$. Indeed, let $$\gamma :\mathbb R \to X$$ be the geodesic starting at $$x$$ with velocity $$u$$, then

\begin{aligned} L\left(\gamma|_{[s,1]}\right) = \int_s^1 |\dot\gamma (t)| dt = |u|\int_s^1 dt = |1-s||u| \end{aligned}

Since the radial geodesic $$\exp_x(su)$$ is the unique minimizing geodesic from $$\exp_x(u)$$ to $$\exp_x(su)$$ it follows

$$d(\exp_x(su) , \exp_x(u)) = |1-s||u|$$

Decomposing $$v = \alpha u + z$$, where $$z$$ is the tangent vector to the geodesic sphere $$\partial_s B(x)$$. By Gauss Lemma this is an orthogonal decomposition, plus there is a curve $$\beta : \mathbb R \to X$$ starting at $$x$$ with velocity $$z$$ such that $$f (\beta(t),y)$$ is constant.

Now define $$\hat f : X \to \mathbb R$$, by $$x \mapsto f(x, \exp_x(u))$$ and $$l : \mathbb \to X$$, by $$t \mapsto \exp_x(tv)$$ then

\begin{aligned} \frac{d}{dt}\bigg|_{t=0}(f \circ l)(t) & = df_0(l(0))\cdot l'(0) = df_0(x)(v) \\&= df_0 (x)(\alpha u + z)\\&=\alpha df_0(x)(u) + \underbrace{df_0(x)(z)}_{0}\\&= \alpha \frac{d}{ds}\bigg|_{s=0} \left(-\frac{1}{2}d(\exp_x(su), \exp_x(u))\right) \\&= \alpha \frac{d}{ds}\bigg|_{s=0} \left(-\frac{1}{2} |1-s|^2|u|\right) \\&= \alpha |u|^2\\&=g_x(u,v) \end{aligned} as we wanted.

Evaluating $$\tilde g_x (v) = g_x(v , \cdot)$$ and $$d_xf (\cdot)$$ at $$v$$, by the claim we get

$$|v|^2 = d_xf (v) = \frac{d}{dt}\bigg|_{t=0} \left(-\frac{1}{2} d(\exp_x(tv), \exp_x (u))^2\right) = g_x (u,v)$$

For any $$v' \perp v$$ we have that

$$0 = d_xf (v') = g_x (u,v')$$ Therefore, $$u=v$$ and $$y = \exp_x (v)$$. To solve $$(\star \star)$$ let $$W = \frac{d}{dt}\Big|_{t=1} \exp_x (tv)$$ and fix any $$w' \perp w$$. Again by the Gauss Lemma $$d_f (w') =0$$. Thus $$w = k W$$. Since geodesics have constant velocity $$|W|^2 = |v|^2$$. Hence the left hand side of $$(\star \star)$$ is

$$g_x(w,W) = k |W|^2 = k |v|^2$$ while the right hand side is

\begin{aligned} -d_yf (W) &= \frac{d}{ds}\bigg|_{s=0} \left(\frac{1}{2}d(x , \exp_x (1+s) (v))^2\right) \\&= \frac{d}{ds}\bigg|_{s=0} \left(\frac{1}{2}d(\exp_x (0v), \exp_x (1+s) (v))^2\right) \\&\underset{(*)}= v\frac{d}{ds}\bigg|_{s=0} \left(\frac{1}{2}(1+s)^2|v|^2\right)\\&= |v|^2 \end{aligned} where we used the same first argument of the claim in $$(*)$$. Thus we have

$$k |v|^2 = |v|^2 \implies k =1$$ and $$w = W$$ as required.