So the equation is:
$$ x^2 + 2y^2 - 6x + 4y + 7 = 0 $$
Find the coordinates of the center, the foci, and the vertex or vertices.
What I did was put the equation in the form: $$ \frac{(x-3)^2}{4}+ \frac{(y+1)^2}{4} = 1 $$
Now based on that, I said the center is at $(3,-1)$, the foci are at $\approx \pm 2.45$ (since $c = \sqrt {a^2 + b^2}$). So the coordinates of that are $(3+2.45,-1)$ and $(3-2.45,-1)$ and the vertices are $(1,-1)$ and $(5,-1)$. I also went ahead and found the asymptote, which is just done by setting the equation to $0$, correct?
The equation should be $$\frac{(x-3)^2}{4}+\frac{(y+1)^2}{2}=1.$$ You've correctly identified the center and vertices. The focal length should be $\sqrt{a^2-b^2}$, not $\sqrt{a^2+b^2}$. Ellipses don't have asymptotes, you're thinking of hyperbolae.