I would like to prove the following statement:
Suppose $\lambda$ is a $1$-form on a manifold $M$ such that $\omega = d \lambda$ is a symplectic form, i.e. non-degenerate, and if we denote $X_{\lambda}$ the vector field which satisfies $\omega(X_{\lambda}, \cdot) = \lambda$, suppose $X_{\lambda}(f) > -1$, where $f$ is a fixed smooth function on $M$. Show that $d(e^f \lambda)$ is also a symplectic form.
Here's what I have:
$$d(e^f \lambda) = e^f(df \wedge \lambda + \omega).$$
My first idea was to show that $(d(e^f \lambda))^{\wedge n}$ is nowhere vanishing. By the binomial formula, I got
$$(d(e^f \lambda))^{\wedge n} = e^{nf} [(n \space df \wedge \lambda+ \omega) \wedge \omega^{\wedge (n-1)}].$$
If $\lambda_{p} = 0$, then at this point the $n$th power is just $e^f \omega^{\wedge n}$, which does not vanish at $p$.
On the other hand, if $\lambda_{p} \neq 0$, then $n \space df \wedge \lambda + \omega \neq 0$ at that point: this is because, for any vector $Y$ such that $\lambda(Y)>0$ we have:
$$(n \space df \wedge \lambda)(X_{\lambda}, Y) = n \space df(X_{\lambda}) \lambda(Y) = n \space X_{\lambda}(f) \lambda(Y) > - \lambda(Y) = - \omega(X_{\lambda}, Y).$$
Now, I'd like to have a result of the type "if two forms are non-zero then their wedge is also non-zero" but I don't think this is true, so I can't conclude. Is there a way to complete this proof?
EDIT: I forgot an $n$ term, so everything I wrote up to now is unfortunately incorrect. I'll see if I can obtain any other kind of partial progress.
I think I figured it out, although I'd also appreciate if anyone were to read the answer and point out any potential mistakes or unclear parts.
I'll prove by hand that $df \wedge \lambda + \omega$ is non-degenerate:
First, if at a point $p \in M$, $\lambda_{p} = 0$, then $d(e^f \lambda)_{p} = e^{f(p)}\omega_{p}$ which is non-degenerate.
Now, let $\lambda_{p} \neq 0$, which means $X_{\lambda}(p) \neq 0$.
Let $X \neq 0$ be an arbitrary vector.
$$(df \wedge \lambda + \omega)(X, X_{\lambda}) = \lambda(X_{\lambda})df(X) -\lambda(X)df(X_{\lambda}) - \lambda(X) = \omega(X, X_{\lambda})[1+df(\lambda)] \neq 0,$$
since $\lambda(X_{\lambda}) = 0$ and $df(\lambda)>-1$.
This means that $df(X) \lambda(Y) = -\omega(X, Y) = \omega(Y, X)$ for all $Y \neq 0$.
If $df(X) = 0$, then $\omega(X, \cdot) = 0$, which is a contradiction by the non-degeneracy of $\omega$.
Therefore by rescaling we may suppose $df(X) = 1$. Then $\lambda(Y) = \omega(-X, Y)$, but $\lambda(Y) = \omega(X_{\lambda}, Y)$, so we have $X = -X_{\lambda}$, since $Y$ was arbitrary. However, then $1 = df(X) = -df(X_{\lambda})$, which is a contradiction since $df(X_{\lambda}) > -1$.