Transfer function of Op-Amp-Twin-T Filter
The way I approach these is using a known opamp configuration and solve the system in sections using Z impedance's for components: 


Source: http://sim.okawa-denshi.jp/en/opampkeisan.htm

The equation for this is: 

\${\Large \frac{V_{out}}{V_{in}}=\frac{Z_{2}}{Z_{1}}}\$

Which I'll call Z7 and Z8

\${\Large \frac{V_{out}}{V_{in}}=\frac{Z_{8}}{Z_{7}}}\$

Now solving for a single T filter:







simulate this circuit &ndash; Schematic created using CircuitLab

you get:

\${\Large \frac{V_{a}}{V_{b}}=\frac{Z_{3}^2}{Z_{3}^2+Z_{1}Z_{3}-Z_{1}Z_{2}}}\$

Now if we duplicate the section with Z4,Z5 and Z6 (Z4 in the place of Z1 in the above pic, Z5 for Z2, Z6 for Z3), since Va and Vb would be the same for both top and bottom t filters we can treat them as parallel impedances

Here are the two parallel impedance's:







simulate this circuit

\${\Large Z_{top}=\frac{Z_{3}^2}{Z_{3}^2+Z_{1}Z_{3}-Z_{1}Z_{2}}}\$
\${\Large Z_{bot}=\frac{Z_{6}^2}{Z_{6}^2+Z_{4}Z_{6}-Z_{4}Z_{5}}}\$

and the parallel impedance's (with the negative terminal of the op amp being Va from above and Vb being Vout:

\${\Large \frac{V_{negterminal}}{V_{out}}=\frac{1}{\frac{1}{Z_{top}}+\frac{1}{Z_{bot}}}}=Z_{7}\$

\${\Large \frac{V_{negterminal}}{V_{out}}=\frac{1}{\frac{Z_{1}}{Z_{3}} +\frac{Z_{4}}{Z_{6}} - \frac{Z_{1}Z_{2}}{Z_{3}^2} - \frac{Z_{4}Z_{5}}{Z_{6}^2} + 2 }=Z_{7}}\$

(confusing yet?) back to the original transfer function:
\${\Large \frac{V_{out}}{V_{in}}=\frac{Z_{8}}{Z_{7}}= \frac{Z_{filt}}{Z_{7}}}\$

where 
\${\Large Z_{7}= \frac{1}{\frac{Z_{1}}{Z_{3}} +\frac{Z_{4}}{Z_{6}} - \frac{Z_{1}Z_{2}}{Z_{3}^2} - \frac{Z_{4}Z_{5}}{Z_{6}^2} + 2 }}\$

and the whole enchilada: 

\${\Large \frac{V_{out}}{V_{in}}=\frac{Z_{8}}{Z_{7}}= Z_{filt}*(\frac{Z_{1}}{Z_{3}} +\frac{Z_{4}}{Z_{6}} - \frac{Z_{1}Z_{2}}{Z_{3}^2} - \frac{Z_{4}Z_{5}}{Z_{6}^2} + 2 )}\$

Now substitute in all of the impedance's for their corresponding values and you get your final equation

\${\Large Z_{1}=\frac{1}{C_1 s} }\$
\${\Large Z_{2}=\frac{1}{C_2 s} }\$
\${\Large Z_{3}=R_3 }\$
\${\Large Z_{4}=R_1 }\$
\${\Large Z_{5}=R_2 }\$
\${\Large Z_{6}=\frac{1}{C_3 s} }\$ 

\${\Large Z_{8}=R_{in} }\$   

Now for a few notes, I double checked this but have been liable to make mistakes. If you follow the process that is not wrong, whith so many variables it's hard to keep track of them all. 

Second thing: I eliminated V3 because it doesn't make sense, if you set the negative terminal to V3 (by having a voltage source on an opamp terminal) then it sets the negative node to that voltage and turns the whole circuit into a comparator based off of Vin3 and ground. This effectively shuts down any feedback you would have.

Third thing: I neglected Cin because I didn't understand your diagram, it should be easily convertible for solving Z8 for a high pass filter or low pass filter depending on your configuration. I just used Rin instead.