Frequency response - finding transfer function
You're on the right track but you need to get your transfer function into standard form.

Gathering terms and multiplying through, we get:

\$\dfrac{V_o}{V_i} = H(j\omega) = \dfrac{1 - j\omega R_6 C_1}{1 + j\omega R_6 C_1}\$

This is an all-pass filter with a magnitude response of 1.  To verify, multiply by the complex conjugate to get \$|H(j \omega)|^2\$:

\$|H(j\omega)|^2 = H H^* = \dfrac{1 - j\omega R_6 C_1}{1 + j\omega R_6 C_1} \dfrac{1 + j\omega R_6 C_1}{1 - j\omega R_6 C_1} = 1\$

For the phase, multiply top and bottom by \$(1 - j\omega R_6 C_1)\$:

\$H(j \omega) = \dfrac{[1 - (\omega R_6 C_1)^2] - j2\omega R_6 C_1}{(1 + \omega R_6 C_1)^2}\$

In this form, the real and imaginary parts are manifest and, for \$\angle H(j \omega) = -90^\circ\$, the real part must be zero.

\$\angle H(j \omega) = \tan^{-1}\frac{-2\omega R_6 C_1}{1 - (\omega R_6 C_1)^2} \$

However, for this particular transfer function (in the first form I wrote), it's easy to see that the numerator and denominator have equal and opposite phase.  Thus, the phase of this transfer function is simply twice the phase of the numerator.  Then, by inspection, we see that \$\angle H(j \omega) = -90^\circ\$ when \$\omega = \frac{1}{R_6 C_1}\$.



(Thanks to Szymon Bęczkowski for the nice plots!)