Generally, a quantum state can be expressed in this form:
$$
|\varphi\rangle = \cos\frac{\theta}{2}|0\rangle + \mathrm{e}^{i\phi}\sin\frac{\theta}{2}|1\rangle
$$
Where $\theta$ and $\phi$ are coordinates on Bloch sphere.
Regarding the particular state in question, we firstly have to get rid of complex amplitude before $|0\rangle$ to have only real number here. We can do that by multiplying whole state by so-called global phase. This multiplication does not change the state as two states which differ in global phase are identical. You can for example check probabilities of $|0\rangle$ and $|1\rangle$ after multiplication. They remain same (for the state in question probability of measuring $|0\rangle$ and $|1\rangle$ in z-basis is $\frac{2}{3}$ and $\frac{1}{3}$, respectively).
Mathematically, the global phase is a complex number with absolute value 1.
In our paritucar case I multiplied $|\varphi\rangle$ with $\frac{1-i}{\sqrt{2}} = \mathrm{e}^{-\frac{\pi}{4}}$ (hence global phase is $-\frac{\pi}{4}$).
The result is
$$
\left| \varphi \right>={\sqrt{\frac{2}{3}}} \left| 0 \right> + \frac{1-i}{\sqrt{2}}\frac{1}{\sqrt{3}} \left| 1\right>
$$
Since $\frac{1-i}{\sqrt{2}} = \mathrm{e}^{-i\frac{\pi}{4}}$, apparently $\phi = -\frac{\pi}{4}$.
Theta can be calculated from $\cos\frac{\theta}{2} = \sqrt{\frac{2}{3}}$. Hence
$$
\theta = 2\arccos\sqrt{\frac{2}{3}} = 1.2310.
$$
We can verify $\theta$ with sine
$$
\theta = 2\arcsin\sqrt{\frac{1}{3}} = 1.2310.
$$
Conclusion: $\theta = 1.2310$ and $\phi = -\frac{\pi}{4}$.
Another numerical example on Bloch sphere coordinates
