Here's a proof along the lines that you are looking for and which is valid only in dimension 2.
Consider a smoothly embedded open ball $B \subset M$. Since $B$ is contractible and the restriction of the symplectic form $\omega$ to $B$ is still closed, this restriction is exact. Thus there exists a 1-form $\lambda$ on $B$ such that $\omega = d \lambda$; by adding to $\lambda$ a closed 1-form and considering if necessary a smaller ball, we can assume without loss of generality that $\lambda$ does not vanish on $B$. From the result mentioned in the question, $\lambda$ can be written as $\lambda = f dg$ for some smooth functions $f, g \in C^{\infty}(B)$. We thus deduce that $\omega = d\lambda = d(fdg) = df \wedge dg$. Hence, considering $\mathbb{R}^2$ equipped with coordinates $(x,y)$ and the standard symplectic form $\omega_0 = dx \wedge dy$, we get that the map $\Phi : B \to \mathbb{R}^2$ given by $\Phi(p) = (f(p), g(p))$ is symplectic, i.e. $\Phi^* \omega_0 = \omega$. It follows that $\Phi$ is an immersion, and it is also a submersion since $B$ and $\mathbb{R}^2$ have the same dimension. This map is thus a local diffeomorphism; Upon shrinking $B$ again if necessary, we can assume that $\Phi$ is a symplectic diffeomorphism onto its image i.e. a Darboux chart.
I am still unsure what are you after; here are some relevant results:
- Suppose that $(M,\omega_0)$ is a compact symplectic manifold. Consider a smooth perturbation of $\omega_0$, i.e. a smooth family of symplectic forms $\omega_t$, $t\in [0,T]$. One question to ask is if there is a smooth family of diffeomorphisms $f_t: M\to M, t\in [0,T]$, such that $f_0=id_M$ and $f_t^*(\omega_t)=\omega_0$. There is an obvious topological obstruction to the existence of such a family, namely, the cohomology classes $[\omega_t]\in H^2(M, {\mathbb R})$ have to be constant (i.e. the same as the one given by $\omega_0$). In other words, for each $t$ there should be a 1-form $\alpha_t\in \Omega^1(M)$ such that $\omega_t- \omega_0= d\alpha_t$.
Now, the relevant theorem is known as Moser's Stability Theorem:
Theorem 1. Assume that in the above setting $[\omega_t]=[\omega_0]$ for all $t$. Then indeed, there is a smooth family of diffeomorphisms $f_t: M\to M, t\in [0,T]$, such that $f_0=id_M$ and $f_t^*(\omega_t)=\omega_0$.
- Moser's theorem generalizes to noncompact manifolds, for instance:
Theorem 2. Suppose that $(M,\omega_t)$ is a symplectic manifold and $\omega_t$ as above is:
a. A compactly supported deformation of $\omega_0$ in the sense that:
There is a compact $K\subset M$ such that $\omega_0=\omega_t$ outside of $K$ for all $t\in [0,T]$, and the compactly supported cohomology class of $(\omega_t-\omega_0), t\in [0,T]$, is zero.
Then there exists a smooth family of diffeomorphisms $f_t: M\to M, t\in [0,T]$, such that $f_0=id_M$ and $f_t^*(\omega_t)=\omega_0$ and, furthermore, $f_t=id, t\in [0,T]$, outside of a compact subset $C\subset M$.
b. In the case when $\omega_0$ is the standard symplectic form on $M={\mathbb R}^{2n}$ one can do a bit better and find a family of diffeomorphisms $f_t: M\to M, t\in [0,T]$ such that $f_0=id_M$ and $f_t^*(\omega_t)=\omega_0$, provided that the difference $\omega_t(x)-\omega_0(x)$ merely decays sufficiently fast (in a suitable sense) as $x\to \infty$.
One can think of Theorem 2 as a version of the Global Darboux Theorem on ${\mathbb R}^{2n}$ for "small perturbations" of the standard symplectic form.
- One can also ask if the Global Darboux Theorem holds for arbitrary symplectic manifolds $(M,\omega)$. One obvious obstruction, of course is that $M=M^{2n}$ is supposed to be diffeomorphic to a domain in ${\mathbb R}^{2n}$. With this restriction, Global Darboux again holds for planar surfaces ($n=1$), due to Greene and Shiohama, generalizing Moser's proof. However, Global Darboux fails in dimensions $\ge 4$ even if $M={\mathbb R}^{2n}$, $n\ge 2$. This was first observed by Gromov (who left a proof as an exercise as he tends to). Explicit examples were found later, for instance, in works by Bates, Peschke and Casals:
Theorem 3. For every $n\ge 2$ there exists a symplectic form $\omega$ on ${\mathbb R}^{2n}$ such that there is no smooth embedding
$f: {\mathbb R}^{2n}\to {\mathbb R}^{2n}$ satisfying
$$
f^*(\omega_0)= \omega,
$$
where $\omega_0$ is the standard symplectic form on ${\mathbb R}^{2n}$.
References:
Larry Bates, George Peschke, A remarkable symplectic structure, J. Differ. Geom. 32, No. 2, 533-538 (1990). ZBL0714.53028.
Roger Casal, Exotic symplectic structures, ZBL07152607.
Robert Greene, Katsuhiro Shiohama, Diffeomorphisms and volume-preserving embeddings of noncompact manifolds, Trans. Am. Math. Soc. 255, 403-414 (1979). ZBL0418.58002.
Jürgen Moser, On the volume elements on a manifold, Trans. Am. Math. Soc. 120, 286-294 (1965). ZBL0141.19407.
Xiudi Tang, "Symplectic Stability and New Symplectic Invariants of Integrable Systems", Ph.D. thesis, 2018.
See also this lecture by Weimin Chen for a self-contained treatment of Moser's theorem.
Best Answer
Darboux's theorem says that any symplectic manifold $(M^{2n}, \omega)$ is locally symplectomorphic to the "trivial" symplectic manifold $( \Bbb R^{2n}, \omega_0)$, where $$\omega_0 = \sum_{j = 1}^n dx^j \wedge dy^j$$ is the standard symplectic form on $\Bbb R^{2n}$. What this means is the following. Given any point $p \in M$, there is a neighborhood $U \subset M$ of $p$ and a diffeomorphism $$\psi: U \longrightarrow \Bbb R^{2n}$$ such that $\psi^\ast \omega_0 = \omega$ on $U$, i.e. $\psi$ is a symplectomorphism. Note that we do not throw away the symplectic structure in this notion of equivalence; without the symplectic structure there is no symplectic geometry to speak of in the first place. "Equivalence" between two symplectic manifolds means that they are symplectomorphic.
A symplectic manifold $(M, \omega)$ is different from a Riemannian manifold $(N, g)$. A symplectic form $\omega$ is a nondegenerate $2$-form, i.e. a nondegenerate antisymmetric rank $2$ tensor. A Riemannian metric $g$, on the other hand, is a nondegenerate symmetric rank $2$ tensor. So right away we see that symplectic manifolds and Riemannian manifolds are fairly different objects. But this is not the reason we say that Riemannian geometry and symplectic geometry are totally different subjects. Darboux's theorem implies that there are no local invariants of symplectic manifolds; any such invariant would also be a symplectomorphism invariant of $(\Bbb R^{2n}, \omega_0)$, and hence we wouldn't be able to discern anything special about a symplectic manifold $(M, \omega)$ by inspecting it locally. This is in stark contrast to Riemannian geometry, where the curvature of a Riemannian manifold is a local object.
Therefore the reason we say symplectic geometry is a totally different subject than Riemannian geometry is that symplectic manifolds have no local symplectomorphism invariants, while local isometry invariants of Riemannian manifolds such as curvature are quite central to the field of Riemannian geometry.
Added: The proof of Darboux's theorem relies on the fact that a symplectic form $\omega$ is closed, i.e. $d\omega = 0$. If $\omega$ is not closed (i.e. we have an almost symplectic manifold) then we do not necessarily have Darboux's theorem. So while Riemannian geometry is in some sense about nondegenerate symmetric bilinear forms on tangent spaces, symplectic geometry is about nondegenerate antisymmetric bilinear forms on tangent spaces with an extra condition described by the differential equation $d\omega = 0$. The extra condition is what makes the theory qualitatively different.
We can impose a differential equation for the metric on a Riemannian manifold to get a "version of Darboux's theorem" for Riemannian geometry. The analogous condition in Riemannian geometry is having a metric $g$ whose Riemann curvature tensor is everywhere zero (i.e. a flat metric); such a manifold is locally isometric to $\Bbb R^n$ with the standard metric.