I've got a problem (da Silva homework 3, question 3) that I have rewritten as the following:
Let $g:M\to M$ be a symplectomorphism such that it takes $(q,p)\to (q',p')= (f(q,p), h(q,p))$.
I want to show that $\partial f/\partial p=0$ and $\partial^2h/\partial p^2=0$.
I have by the assumption that $g^{*}\alpha=\alpha$ that $p'dq'=pdq$ i.e. $h(q,p)dq'=pdq$ [$\alpha$ is the tautological 1-form].
I have that $dq'=\frac{\partial f}{\partial p}dp+\frac{\partial f}{\partial q}dp.$ Therefore $h(q,p)dq'=pdq$ implies that
$$h\left(\frac{\partial f}{\partial p}dp+\frac{\partial f}{\partial q}dp\right)=pdq$$ If it is the case that $h(q,p)=0$, then we have that $h$ is the zero function so $h(q,\lambda p)=h(q,p)=0$ and $\frac{\partial^2 h}{\partial q^2}=0$.
So I'm stuck when $h\neq 0$. This tells us that $\frac{\partial f}{\partial p}=0$. Then I know $g=g(q)$ and it remains to show that $\frac{\partial^2h}{\partial q^2}=0$.
I tried doing IBP by showing that $$\int h(q,p)\frac{\partial f}{\partial q}dq=\int pdq$$ implies $$ h(q,p)f(q,p)-\int \frac{\partial h}{\partial q}f(q,p)dq=qp-\int qdp$$.
I'm not sure what to do at this point or if this is even a reasonable approach. I also wonder if there is a way to do IBP that can swap the $dq$ and $dp$? By this I mean if there is a way to write $\int \frac{\partial h}{\partial q}f(q,p)dq$ as something with $\int \text{blah} dp$?
Since $q'=f$ your $dq'$ should be $$ dq'=\frac{\partial\color{red}f}{\partial p}\,dp+\frac{\partial\color{red}f}{\partial q}\,d\color{red}q\,. $$ From $p'\,dq'=h\,dq=p\,dq$ it then follows that $$\tag{1} \frac{\partial\color{red} f}{\partial p}\equiv 0\,. $$
Likewise, from $$ dp'=\frac{\partial h}{\partial p}\,dp+\frac{\partial h}{\partial q}\,dq $$ and $dp'\wedge dq'=dp\wedge dq$ it follows that $$ dp\wedge dq=\frac{\partial h}{\partial p}\frac{\partial f}{\partial q}\,dp\wedge dq+\underbrace{\frac{\partial h}{\partial q}\frac{\partial f}{\partial p}\,dq\wedge dp}_{0}\,. $$ Therefore $$\tag{2} \frac{\partial h}{\partial p}\frac{\partial f}{\partial q}\equiv 1\,. $$ Taking a derivative gives $$ \frac{\partial^2 h}{\partial p^2}\frac{\partial f}{\partial q}+\frac{\partial h}{\partial p}\!\!\!\!\!\underbrace{\frac{\partial^2 f}{\partial p\,\partial q}}_{=0\text{ because of }(1)}\equiv 0\,. $$ By (2) it is not possible that $\frac{\partial f}{\partial q}$ vanishes anywhere. Therefore $$ \frac{\partial^2 h}{\partial p^2}\equiv 0\,. $$