De Sauty bridge with RC parallels
Your first equation is of course correct:

$$Z_x=Z_c\frac{R_1}{R_2}$$
Now it's convenient to invert this equation:

$$\frac{1}{Z_x}=\frac{1}{Z_c}\frac{R_2}{R_1}$$

Using \$1/Z_x=j\omega C_x+1/R_x\$ and \$1/Z_c=j\omega C_c+1/R_c\$ we obtain

$$j\omega C_x+1/R_x = (j\omega C_c+1/R_c)\frac{R_2}{R_1}$$
Comparing real and imaginary parts we get

$$C_x=C_c\frac{R_2}{R_1}\text{ and }R_x=R_c\frac{R_1}{R_2}$$