Phase and group velocity calculation

I have a wave described by $$A \sin(k_x x) \exp(i(k_y y− ωt))$$. This wave is propagating in $$+y$$ direction. However its angular frequency is given by $$k_x^2 + k_y^2 = (w/c)^2$$. If I want to calculate phase and group velocity does it mean that I only use $$k_y$$, because the wave is travelling in that direction? For example $$v_{\rm p} = \omega/k_y$$ and $$v_{\rm g} = d\omega/dk_y$$?

Your wave has the form $$f(x) e^{i(k_y y - \omega t)}$$ where $$f(x)$$ is some function of $$x$$; in this case $$A \sin(k_x x)$$. So you have a wave travelling in $$+y$$ direction as you say, and all that $$f(x)$$ is doing is saying that the amplitude of the wave is a function of $$x$$. I hope this is enough to make it clear to you that the phase velocity is indeed $$v_{\rm p} = \omega/k_y$$ since this is the speed at which the wavefronts are moving. (A wavefront being a locus of points at constant phase of the wave).
Once we have the phase velocity at each $$\omega$$ and $$k_y$$, the group velocity follows. To see this, consider a pair of waves of the type under discussion, one at $$k_x, k_y$$ and one at $$k_x + dk_x, \; k_y + dk_y$$. Add the two waves together, and you should find you have a fast oscillation and a beat pattern envelope. That envelope moves at speed $$d \omega / d k_y$$. Using the dispersion relation you then get $$\frac{d \omega}{d k_y} = c \frac{d}{d k_y} \left(k_x^2 + k_y^2 \right)^{1/2}$$ and to carry out the differentiation you would need to know how $$k_x$$ and $$k_y$$ are related for the waves you are considering.