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?
More generally, let there be given a Poisson manifold $(M,\pi)$, where $$\pi ~=~ \frac{1}{2} \pi^{IJ} \frac{\partial}{\partial z^I} \wedge \frac{\partial}{\partial z^J} $$ is a Poisson bi-vector, and $$\{ f, g\}_{PB}~=~\frac{\partial f}{\partial z^J}\pi^{IJ}\frac{\partial g}{\partial z^J} $$ is the corresponding Poisson bracket. Let the Hamiltonian $H$ be a globally defined function on $M$. Then Hamilton's equations read $$ \dot{z}^{I}~=~\{ z^I, H\}_{PB}, $$ i.e. time-evolution is given by (minus) the Hamiltonian vector field $$ X_H~=~\{H,\cdot\}_{PB}. $$
If the Poisson structure is invertible, then $M$ is a symplectic manifold with symplectic 2-form $$\omega ~=~\frac{1}{2} \omega_{IJ}~ \mathrm{d}z^I \wedge \mathrm{d}z^J,$$ where $\omega_{IJ}$ is the inverse matrix: $$ \pi^{IJ}\omega_{JK}~=~\delta^I_K. $$
In canonical/Darboux coordinates $$ (z^1, \ldots, z^{2n})~=~(q^1, \ldots, q^n,p_1,\ldots, p_n) ,$$ the above construction reduces to the standard Poisson bi-vector $$\pi~=~\frac{\partial}{\partial q^i} \wedge \frac{\partial}{\partial p_i},$$ and the standard symplectic 2-form $$\omega ~=~ \mathrm{d}p_i \wedge \mathrm{d}q^i.$$
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}
