So I am struggling with this question: Is following function a wave function? If yes, what is the wave propagation velocity? E, $\alpha$ and $\beta$ are constants.
$$\phi=Ee^{\alpha x^2+\beta^2\alpha t^2-2\beta\alpha xt}$$
So I know I can write this as follows to get a wave equation of the form $Ae^{i(kx-\omega t)}$:
$$\Rightarrow\phi=Ee^{{-(i(\sqrt\alpha x+\beta\sqrt\alpha t))}^2}$$
Now normally the velocity of a regular wave equation is $v\ =\ \frac{\omega}{k}$ . Is this still the case in the upper equation with that second power? Or is this not a wave equation anymore?
The wave equation is a partial differential equation. When we say, "is $\phi$ a wave equation" we mean "is it a solution to the PDE?"
$$ \Delta \phi = \frac{1}{v^2} \dfrac{\text{d}^2\phi}{\text{d}t^2} $$
is the basic wave equation where $\Delta \phi$ denotes the Laplacian (just the second partial derivative for $x$ in your case). Observe that
$$ \Delta \phi = E(2\alpha x-2\beta\alpha t)^2e^{\alpha x^2+\beta^2\alpha t^2-2\beta\alpha xt} + E(2\alpha)e^{\alpha x^2+\beta^2\alpha t^2-2\beta\alpha xt} $$
and
$$ \dfrac{\text{d}^2\phi}{\text{d}t^2} =(2\beta^2\alpha t-2\beta\alpha x)^2Ee^{\alpha x^2+\beta^2\alpha t^2-2\beta\alpha xt} + (2\beta^2\alpha)Ee^{\alpha x^2+\beta^2\alpha t^2-2\beta\alpha xt} \text{ .}$$
The rest is just algebra so I'll leave you to work out the details. Keep in mind that $1/v^2$ is just a constant. If you can write $ \Delta \phi $ in such a way that it looks like a constant multiple of $ \dfrac{\text{d}^2\phi}{\text{d}t^2} $ then you will have shown that $\phi$ is a wave equation and that the constant is in fact $1/v^2$.