Firstly, note that the state is seperable (not entangled):

\begin{equation}
|\psi\rangle = \frac{1}{\sqrt{3}} |00\rangle + \frac{1}{\sqrt{3}} |01\rangle + \frac{1}{\sqrt{6}} |10\rangle + \frac{1}{\sqrt{6}} |11\rangle= \\
= \left(\frac{\sqrt{2}}{\sqrt{3}}|0\rangle + \frac{1}{\sqrt{3}}|1\rangle \right) \left(\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle \right) = |\psi_1 \rangle |\psi_2 \rangle
\end{equation}

where $|\psi_1 \rangle = \left(\frac{\sqrt{2}}{\sqrt{3}}|0\rangle + \frac{1}{\sqrt{3}}|1\rangle \right)$ and $|\psi_2 \rangle= \left(\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle \right)$ correspond to the first and the second qubit states. This $|\psi\rangle = |\psi_1 \rangle |\psi_2 \rangle$ is not possible if we have entangled states. The consequence of this is that after measuring $|0\rangle$ for the first qubit (in the example the measurement is done **for the first qubit**), the state of the first qubit becomes $|\psi_1 \rangle \rightarrow |0\rangle$ (the second qubit's state will not be changed) and the combined state:

$$|\psi\rangle= |0\rangle \left(\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|01\rangle \right) = \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|01\rangle$$

**The explanation with the projective measurements:**

We should renormalize the state after the measurement (after measurement the probabilities of the state should sum to $1$). If we apply the projective measurement, we should calculate the probability of measuring $m$ outcome:

$$p(m) = \langle \psi | P_m | \psi \rangle$$

where $P_m$ is is the projector onto the eigenspace of $M$ with eigenvalue $m$, $M$ is a Hermitian operator/observable, that describes the measuremet. Then the state after the measurement outcome $m$ will be equal to:

$$\frac{P_m |\psi \rangle}{\sqrt{p_m}}$$

The division to the $\sqrt{p_m}$ is for the renormalization of the state after the action of the projector. A more rigorous definition can be found in M. Nielsen and I. Chuang's textbook page 87.

In the case of the question the observable is $M = Z\otimes I$, the projector to the $|0\rangle$ state of the first qubit is $P_{+1} = |0\rangle \langle 0| \otimes I$, the $I$ means that we are not touching the second qubit, $m$ eigenvalue is $+1$. Then the probability of measuring the first qubit $|0\rangle$:

$$p_{+1} = \langle \psi | P_{+1} | \psi \rangle = \frac{2}{3}$$

The resulting state:

$$\frac{P_{+1} |\psi \rangle}{\sqrt{p_{+1}}} = \frac{\frac{1}{\sqrt{3}}|00\rangle + \frac{1}{\sqrt{3}}|01\rangle }{\sqrt{\frac{2}{3}}} = \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|01\rangle$$

1The first qubit is measured; and since it is measured $|0\rangle$, the post-measurement state of the second qubit is $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$. I don't quite understand what exactly is asked. – kludg – 2020-05-16T06:24:37.787