Chern class on a symplectic manifold Let $(X,\omega)$ be a closed symplectic manifold.  Can we always write $c_1(TX) = [ f \omega ]$
for some function $f: X \to \mathbb{R}$?
 A: Your question is correspond to weak Kahler-Einstein metric(I know for Kahler case). Let me explain in more details
The Hermitian metric $H$ along the fibers of the holomorphic vector bundle $E$ over a K\"ahler manifold $(M,\omega)$ is weak Hermitian-Einstein if $$\sqrt[]{-1}\Lambda_\omega F_H =
\phi I$$ at every point of $M$, where $\gamma$ is a function on $M$ and $I$ is the identity
endomorphism of $E$ and $F_H$ is the Chern curvature form of $(E, h)$. Here $\Lambda_\omega$ is the contraction with $\omega$ which means 
$$F_H\wedge\omega^{n-1}=(n-1)!\Lambda_\omega F_H\frac{\omega^n}{n!}$$
If $\Phi$ is constant $\lambda$ then we call it Hermitian-Einstein metric. As additional remark of the sign of $\lambda$,
If the metric $h$ admit hermitian-Einstein metric $Λ_ωF_h=λI $, and if $λ<0$ there is no holomorphic section, and if $λ=0$ then all holomorphic sections are parallel. Since $$\Delta |s|^2=|\nabla s|^2-<(\Lambda_\omega F_h)s,s>\geq 0$$
where $s$ is a section and hence $|s|$ is bounded. But maximum principle for subharmonic functions, so $|∇s|^2=λ|s|^2$. Hence if $λ<0$ there is no holomorphic section, and if $λ=0$ then all holomorphic sections are parallel.
Now take $E=TM$, then we can define weak Kahler-Einstein metric as $Ric(\omega)=\phi \omega$ and hence $c_1(TM)=[Ric(\omega)]=[\phi \omega]$
When $\phi$ is globally constant, then we call it Kahler-Einstein metric and when it is positive we call $M$ is Fano and when it is negative we call $M$ is of general type and it is zero $M$ is called Calabi-Yau variety
From Kobayashi book (complex vector bundle ) we have the following theorem for more general setting of your question
Theorem: An Hermitian vector bundle $(E, h)$ over a Riemann surface
$(M, g)$ satisfies the weak Einstein condition if and only if it is projectively flat.
Note that in general one direction always works
Theorem: Let $(M, g)$ is a Kahler manifold, Let $(E, h)$ be a projectively flat vector bundle over $M$.
Then, for any Hermitian metric $g$ on $M$,$(E, h)$ satisfies the weak Einstein
condition.
We have also the following theorem which tell us that any weak Hermitian-Einstein metric after conformal change is Hermitian-Einstein metric
Theorem: If an Hermitian vector bundle $(E, h)$ over a compact Kahler
manifold $(M, g)$ satisfies the weak Einstein condition with factor $φ$, there is a conformal change $h −→ h′ = ah$ such that $(E, h′ )$ satisfies the Einstein condition with a constant factor $c$. Such a conformal change is unique up to a
homothety. Now you can take $E=TM$ and this tells us that any weak Kahler-Einstein there is a conformal change  $h −→ h′ = ah$ such that $(TM, h′ )$ is Kahler-Einstein metric
A: Take a product of spheres, and let $\omega$ be half the area form on the first factor plus twice that on the second factor. The Chern classes of the tangent bundle are clearly invariant under switching the factors, since we can deform the symplectic structure into the standard Kaehler structure, and the Chern classes are integer classes. As divertietti points out, the function $f$ in your description must be constant, so doesn't give the correct symplectic form.
A: The answer is no.
First of all, if you want $f\omega$ to define a cohomology class, you should ask that $d(f\omega)=0$, and this is equivalent to $df\wedge\omega=0$, since $d\omega=0$. Using Darboux coordinates, it is straightforward to see that this implies that $f$ must be (locally) constant (see e.g. this answer).
Now, take your $X$ to be a connected compact Kähler manifold, which is of course symplectic, once you fix a Kähler form $\omega$. Thus, you are asking whether you can find a real constant $\lambda$ such that $c_1(X)=[\lambda\omega]$. This implies in particular that the sign of $c_1(X)$ must be definite. To get a counterexample it is therefore sufficient to take any compact Kähler manifold whose canonical bundle is not positive, nor negative, nor zero.
