You can make use of the Schmidt-decomposition-based method discussed generally in Section VII of this paper. For the particular case of the preparation of a three-qubit state, such as the one considered in this question, you may refer to Appendix F of the same paper, which is what I shall follow below.

First, we cast the $8$-dimensional vector corresponding to the three-qubit state,

$|\psi \rangle = (0, 0, \frac{1}{2}, 0, 0, \frac{\sqrt{3}}{2}, 0, 0)^{T}$,

into a $4 \times 2$ matrix

$M' = \begin{pmatrix} 0 & 0 \\ \frac{1}{2} & 0 \\ 0 & \frac{\sqrt{3}}{2} \\ 0 & 0 \end{pmatrix}$,

which effectively defines an asymmetric bipartition with two qubits on one side and one qubit on the other. Now we can perform the singular value decomposition of matrix $M' = U' S' V'^{\dagger}$, with

$U' = \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}, \quad S' = \begin{pmatrix} \frac{\sqrt{3}}{2} & 0 \\ 0 & \frac{1}{2} \\ 0 & 0 \\ 0 & 0 \end{pmatrix} \equiv \begin{pmatrix} s'_0 & 0 \\ 0 & s'_1 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}, \quad V'^{\dagger} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$

Finally, we just have to implement the quantum circuit shown in Fig. 10 of Appendix F of the reference paper, which requires expressing $U'$ and $V'$ as two-qubit and single-qubit subcircuits, as well as finding a single-qubit subcircuit $B'$ that prepares the state $s_0 |0\rangle + s_1 |1\rangle \equiv \frac{\sqrt{3}}{2} |0\rangle + \frac{1}{2} |1\rangle$ starting from the fiducial state $|0\rangle$.

• $B' = R_y(\frac{\pi}{3})$.
• $V' = X$.
• $U' = C_{0}NOT_{\textrm{least}, \textrm{most}} (Z \otimes \mathbf{1})$.

The notation $C_{0}NOT_{\textrm{least}, \textrm{most}}$ should be clarified: It is a CNOT gate with the least significant qubit as the control-qubit, the most significant qubit as the target-qubit, and the $NOT$ gate only being applied to the target-qubit when the control-qubit is in state $|0 \rangle$. The $Z$ gate prior to this CNOT is applied to the most significant qubit of the pair.

Putting all the pieces together, the quantum circuit that prepares state $|\phi \rangle$ is shown below. The qubits are ordered from least to most significant from top to bottom.

More generally, this method allows to prepare any three-qubit state with at most $4$ CNOTs, since any two-qubit operation $U'$ can be implemented by a quantum circuit with at most $3$ CNOTs.

You can make use of the Schmidt-decomposition-based method discussed generally in Section VII of this paper. For the particular case of the preparation of a three-qubit state, such as the one considered in this question, you may refer to Appendix F of the same paper, which is what I shall follow below.

First, we cast the $8$-dimensional vector corresponding to the three-qubit state,

$|\psi \rangle = (0, 0, \frac{1}{2}, 0, 0, \frac{\sqrt{3}}{2}, 0, 0)^{T}$,

into a $4 \times 2$ matrix

$M' = \begin{pmatrix} 0 & 0 \\ \frac{1}{2} & 0 \\ 0 & \frac{\sqrt{3}}{2} \\ 0 & 0 \end{pmatrix}$,

which effectively defines an asymmetric bipartition with two qubits on one side and one qubit on the other. Now we can perform the singular value decomposition of matrix $M' = U' S' V'^{\dagger}$, with

$U' = \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}, \quad S' = \begin{pmatrix} \frac{\sqrt{3}}{2} & 0 \\ 0 & \frac{1}{2} \\ 0 & 0 \\ 0 & 0 \end{pmatrix} \equiv \begin{pmatrix} s'_0 & 0 \\ 0 & s'_1 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}, \quad V'^{\dagger} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$

Finally, we just have to implement the quantum circuit shown in Fig. 10 of Appendix F of the reference paper, which requires expressing $U'$ and $V'$ as two-qubit and single-qubit subcircuits, as well as finding a single-qubit subcircuit $B'$ that prepares the state $s_0 |0\rangle + s_1 |1\rangle \equiv \frac{\sqrt{3}}{2} |0\rangle + \frac{1}{2} |1\rangle$ starting from the fiducial state $|0\rangle$.

• $B' = R_y(\frac{\pi}{3})$.
• $V' = X$.
• $U' = C_{0}NOT_{\textrm{least}, \textrm{most}} (Z \otimes \mathbf{1})$.

The notation $C_{0}NOT_{\textrm{least}, \textrm{most}}$ should be clarified: It is a CNOT gate with the least significant qubit as the control-qubit, the most significant qubit as the target-qubit, and the NOT gate only being applied to the target-qubit when the control-qubit is in state $|0 \rangle$. The $Z$ gate prior to this CNOT is applied to the most significant qubit of the pair.

Putting all the pieces together, the quantum circuit that prepares state $|\psi \rangle$ is shown below. The qubits are ordered from least to most significant from top to bottom.

More generally, this method allows to prepare any three-qubit state with at most $4$ CNOTs, since any two-qubit operation $U'$ can be implemented by a quantum circuit with at most $3$ CNOTs.