How do we know \$R_x=\frac{R_2R_3}{R_1}\$,and \$L_x=C_1R_2R_3\$ in the maxwell bridge
Start by calculating R1 in parallel with C (product over sum): - 

$$\dfrac{\dfrac{R_1}{j\omega C}}{R_1 + \dfrac{1}{j\omega C}} \Rightarrow  \dfrac{R_1}{1 + j\omega CR_1}$$

Then, multiply top and bottom by: \$1 - j\omega CR_1\$ (this is called using the complex conjugate): -

$$\dfrac{R_1(1 - j\omega CR_1)}{1 + \omega^2 C^2R_1^2}$$

Then multiply by \$R_x + j\omega L\$ to get: -

$$\dfrac{R_1(R_x + j\omega L)(1 - j\omega CR_1)}{1 + \omega^2C^2R_1^2}$$

Notice here that there are complex terms only in the numerator and not the denominator. Notice also that these complex terms cannot equate to anything but zero because ultimately the bridge is in balance and the above formula equals \$R_2R_3\$ (no complex terms). So,

$$j\omega L = j\omega CR_xR_1$$

Hence \$L = CR_xR_1\$ and I think you can see that this agrees with your question's other solution if you accept that it is true: - 

$$R_x=\dfrac{R_2R_3}{R_1}$$

Now I could go on here and prove that but I think you can take it from here. Let me know if you are confused.