Problem: Let $M$ be a manifold, and let $\alpha$ be a closed $1$-form on $M$. Show that there exists a symplectomorphism of $(T^{*} M, \omega_{T^{*} M})$ that maps the zero section to the submanifold $\text{Im}(\alpha) := \left\{ \alpha_x: x \in M \right\} $ of $T^{*} M$.
Attempt: The zero section of $T^{*} M$ is $$M_{0} = \left\{ (x, \xi) \in T^{*} M \mid \xi = 0 \ \text{in} \ T_x^{*} M \right\}. $$
My idea was to define the symplectomorphism $\phi$ as $\phi := \alpha \circ \pi$, where $\pi : T^{*} M \rightarrow M$ is the projection.
I have to show that $\phi$ is a diffeomorphism and that $\phi^{*} (\omega_{T^{*} M}) = \omega_{T^{*} M}$ ? However, this is where I am stuck. Also, I'm not sure where I need the closedness of $\alpha$.
Help is appreciated.
The $\phi$ you defined isn't even a bijective map, since it maps the entire fibre $T_x^*M$ to $\alpha_x$. The appropriate $\phi$ is $\phi = \mathrm{id}_{T^*M}+\alpha\circ\pi$. Then if $\Theta$ is the canonical 1-form on $T^*M$ (defined by $\Theta_{\beta_x}(V_{\beta_x}) := \beta_x(T_{\beta_x}\pi\,(V_{\beta_x}))$ for $V_{\beta_x}\in T_{\beta_x}(T^*M)$), then $$ (\phi^*\Theta)_{\beta_x}(V_{\beta_x}) = \Theta_{\phi(\beta_x)}(T_{\beta_x}\phi\,(V_{\beta_x})) = \phi(\beta_x)(T_{\phi(\beta_x)}\pi\,(T_{\beta_x}\phi\,(V_{\beta_x}))) \\= (\beta_x+\alpha_x)(T_{\beta_x}(\pi\circ\phi)(V_{\beta_x})) = (\beta_x+\alpha_x)(T_{\beta_x}\pi\,(V_{\beta_x}))=(\Theta+\pi^*\alpha)_{\beta_x}(V_{\beta_x}), $$ i.e. $\phi^*\Theta = \Theta+\pi^*\alpha$. Hence $$ \phi^*\omega_{T^*M} = \phi^*(-d\Theta) = -d(\phi^*\Theta) = -d(\Theta+\pi^*\alpha) = \omega_{T^*M}-\pi^*(d\alpha). $$ If $\alpha$ is closed, then $\phi$ is a symplectomorphism.