The second step that I would do is choose an element not from the ones chosen, say $\displaystyle x$. Now we will use the theorem which will make computations a little easier. The number of elements conjugate with $\displaystyle x$ has to be $\displaystyle [G:\text{C}(x)]$. Now $\displaystyle 1,x,x^2,x^3 \in \text{C}(x)$ thus it has

__at least__ four elements, by Lagrange's theorem it can therefore have $\displaystyle 4\mbox{ or } 8$ elements it cannot be $\displaystyle 8$ since $\displaystyle x$ is not in the center thus $\displaystyle \text{C}(x) = \{1,x,x^2,x^3\}$ thus $\displaystyle [G:\text{C}(x)] = 2$ which means the number of elements conjugate with $\displaystyle x$ is just two. Now $\displaystyle yxy^{-1} = x^{3}yy^{-1} = x^3$ thus $\displaystyle x^3$ is conjugate to $\displaystyle x$,

__we can stop here__ by the theorem thus $\displaystyle \{x,x^3\}$ is another conjugacy class.

The third step that I would do is choose an element not from the ones chosen, say $\displaystyle y$. Now we will use the theorem which will make computations a little easier. The number of elements conjugate with $\displaystyle y$ has to be $\displaystyle [G:\text{C}(y)]$. Now $\displaystyle 1,x^2,y\text{C}(y) $ thus it has

__at least__ three elements it cannot have all eight thus by Lagrange's theorem $\displaystyle |\text{C}(y)| = 4$ which implies that $\displaystyle [G:\text{C}(y)] = 2$. Now $\displaystyle xyx^{-1} = x^2y$ thus $\displaystyle x^2y$ is conjugate to $\displaystyle y$,

__we can stop here__ by the theorem thus $\displaystyle \{y,x^2y\}$ is another conjugacy class.