I am not exactly sure whether to ask it on physics SE or math SE, but I think it is suited here
Clearly, the wave is moving in the positive x direction
Using the general wave equation $$y=A\sin(\omega t + kx + \phi _0)$$
At $x=0$ $$\omega t + \phi_0=\frac{\pi}{2}$$ At $t=0$ $$kx+\phi_0=\frac{-\pi}{6}$$
How do I proceed from here?
Firstly, as the wave velocity $=\frac{-w}{k}=v$ in the positive x Direction, let the wave equation be $y=A \sin(wt-kx+\phi)$, to avoid getting negative values later.
Given graph is for particle at $x=0$. So it's motion can be described as $y=A \sin(wt + \phi)$. At $t=0, y=-\frac{A}{2}$. So $\sin(\phi)=-\frac{1}{2}$. Notice that the velocity of the particle is "down"/away from equilibrium position, indicating that the value of $\phi$ is the first solution to the above equation. So $\phi = \frac{7π}{6}$.
To determine $w$ there must be some other information(missing), apart from the given equation, such as intercept make by graph with x axis for the first time etc.
With that information, we can substitute for $y=0$ and solve for $w$. And with this we can get value of k as well, using $\frac{w}{k}=v$ (only the magnitude here, as we have already determined their signs).