Given a symplectic manifold $(M^{2n},\omega)$ is it always true that every top de Rham cohomology class $[\alpha]\in H^{2n}(M)$ is of the form $[\alpha]=[f\omega^n]$ for some smooth function $f\in C^{\infty}(M)$. More restrictedly , is the homomorphism $L^{n}_{[\omega]}:H^0(M)\to H^{2n}(M)$ given by $L^{n}_{[\omega]}([f])=[f\omega^n]$ always surjective?
In the case when $M$ is compact and connected, then this is always true because both $H^0(M)$ and $H^{2n}(M)$ are vector spaces of dimension 1, and $[\omega^n]$ is a non-zero element of $H^{2n}(M)$. Another case is when $\omega$ is exact then $M$ has to be non-compact. Therefore in this case it is true when we assume again that $M$ is connected.
How to prove this result in the general case (especially when $M$ is not necessarily connected).
Even better, this holds at the level of forms and not just at the level of cohomology, i.e. every $2n$-form $\alpha\in\Omega^{2n}(M)$ (note that $2n$-forms are automatically closed) has the form $\alpha=f\omega^n$ for some $f\in C^{\infty}(M)$. Indeed, the $2n$-th exterior power $\bigwedge^{2n}T_pM$ of the tangent space at a point $p\in M$ is one-dimensional, so there is a unique scalar $f(p)\in\mathbb{R}$ s.t. $\alpha\vert_p=f(p)\omega^n\vert_p$. This yields the desired function $f\colon M\rightarrow\mathbb{R}$ and it only remains to argue that $f$ is smooth. However, given a coordinate chart $\varphi=(x^1,\dotsc,x^{2n})\colon U\rightarrow\varphi(U)\subseteq\mathbb{R}^{2n}$, there are smooth functions $g,h\in C^{\infty}(\varphi(U))$ s.t. $\alpha\vert_U=g(x^1,\dotsc,x^{2n})dx^1\wedge\dotsc\wedge dx^{2n}$ and $\omega^n=h(x^1,\dotsc,x^{2n})dx^1\wedge\dotsc\wedge dx^{2n}$, where additionally $h$ is nowhere-vanishing (since $\omega^n$ is a volume form). Thus, $f\circ\varphi^{-1}=\frac{g}{h}$ is smooth. The claim follows.
Note that this really is a statement about oriented manifolds, not symplectic manifolds. The above holds true if $\omega^n$ is replaced with any volume form.