Let $ABP, BCQ, CAR$ be three non-overlapping triangles erected outside of acute triangle $ABC$. Let $M$ be the midpoint of segment $AP$. Given that $\angle PAB = \angle CQB = 45^\circ$, $\angle ABP = \angle QBC = 75^\circ$, $\angle RAC = 105^\circ$, and $RQ^2 = 6CM^2$, we aim to compute $\frac{AC^2}{AR^2}$.

Construct parallelogram $CADP$.

Claim: $\triangle AQR \sim \triangle ADC$.

Proof: Observe that $\triangle BPA \sim \triangle BCQ$, hence $\triangle BAQ \sim \triangle BPC$. Consequently,
$$\frac{AQ}{AD} = \frac{AQ}{CP} = \frac{BP}{BA} = \sqrt{\frac{3}{2}} = \frac{QR}{DC}.$$
Since $\angle RAC = 105^\circ$ and $\angle QAD = \angle CPA + \angle QAP = 180^\circ - \angle (CP, AQ) = 180^\circ - \angle ABP = 105^\circ$, we can use SSA similarity (since $105^\circ > 90^\circ$) to conclude that $\triangle AQR \sim \triangle ADC$.

Thus, it follows that
$$\frac{AC^2}{AR^2} = \frac{2}{3}.$$

The answer is: $\boxed{\frac{2}{3}}$.