Question: Hamilton's equations for a Hamiltonian $H(q,p)$ w.r.t. to a standard symplectic from $\omega = dq \wedge dp$ are $$\dot{q} = \partial H\_{p}, \quad \dot{p} = - \partial H\_{q}$$

How do Hamilton's equations write w.r.t. a nonstandard symplectic form $F(q,p) dq \wedge dp$, where $F(q,p)$ is some smooth function?

Answer: 
1. More generally, let there be given a [Poisson manifold](https://en.wikipedia.org/wiki/Poisson_manifold) $(M,\pi)$, where
$$\pi ~=~ \frac{1}{2} \pi^{IJ} \frac{\partial}{\partial z^I} \wedge \frac{\partial}{\partial z^J} \tag{1}$$
is a Poisson bi-vector, and
$$\{ f, g\}\_{PB}~=~\frac{\partial f}{\partial z^I}\pi^{IJ}\frac{\partial g}{\partial z^J} \tag{2}$$
is the corresponding Poisson bracket. Let the Hamiltonian $H$ be a globally defined function on $M$. Then Hamilton's equations read
$$ \dot{z}^{I}~=~\{ z^I, H\}\_{PB},\tag{3} $$
i.e. time-evolution is given by (minus) the [Hamiltonian vector field](https://en.wikipedia.org/wiki/Hamiltonian_vector_field)
$$ X\_H~=~\{H,\cdot\}\_{PB}.\tag{4} $$
2. If the Poisson structure is invertible, then $M$ is a [symplectic manifold](https://en.wikipedia.org/wiki/Symplectic_manifold)
with symplectic 2-form
$$\omega ~=~\frac{1}{2} \omega\_{IJ}~ \mathrm{d}z^I \wedge \mathrm{d}z^J,\tag{5}$$
where $\omega\_{IJ}$ is the inverse matrix:
$$ \pi^{IJ}\omega\_{JK}~=~\delta^I\_K. \tag{6}$$
3. In canonical/[Darboux](https://en.wikipedia.org/wiki/Darboux%27s_theorem) coordinates
$$ (z^1, \ldots, z^{2n})~=~(q^1, \ldots, q^n,p\_1,\ldots, p\_n) ,\tag{7}$$
the above construction reduces to the standard Poisson bi-vector
$$\pi~=~\frac{\partial}{\partial q^i} \wedge \frac{\partial}{\partial p\_i},\tag{8}$$
and the standard symplectic 2-form
$$\omega ~=~ \mathrm{d}p\_i \wedge \mathrm{d}q^i.\tag{9}$$
A Hamiltonian $H:M\rightarrow \mathbb{R}$ defines a vector field $X\_H$ through the equation
\begin{equation}
 \omega(X\_H,\cdot)=dH.
\end{equation}
For $\omega=F(q,p)dq\wedge dp$ and substituting the components $X\_H=X\_{Hq}\partial\_q+X\_{Hp}\partial\_p$ we get 
\begin{equation}
 F(q,p)(X\_{Hq}dp-X\_{Hp}dq)=(\partial\_qH)dq+(\partial\_pH)dp.
\end{equation}
The integral curves $t\mapsto(q(t),p(t))$ of the vector field $X\_H$ represent the Hamiltonian flow of the system. Therefore, we have
\begin{align}
 \dot{q}=\frac{\partial\_qH}{F(q,p)};\;\;\;
 \dot{p}=-\frac{\partial\_pH}{F(q,p)};
\end{align}