Find the transfer function and determine the type of this filter
I'd use the complex-AC version of Millman's theorem to solve for voltage \$\color{red}{V_X}\$: -

Hence,
$$\color{red}{\boxed{V_X}} = \dfrac{\dfrac{V_I}{R1}+\dfrac{0}{R2 + \frac{1}{sC1}}+\dfrac{V_O}{\frac{1}{sC2}}}{\dfrac{1}{R1} +\dfrac{1}{R2 + \frac{1}{sC1}} +\dfrac{1}{\frac{1}{sC2}}}$$
And, of course for a unity gain non-inverting amplifier: $$V_X\cdot\dfrac{\frac{1}{sC1}}{R2 + \frac{1}{sC1}} = V_O$$
Or, \$\hspace{5.5cm}V_X = V_O\cdot(1+sC1R2)\$
So,
$$V_O\cdot(1+sC1R2) = \dfrac{\dfrac{V_I}{R1}+V_O\cdot sC2}{\dfrac{1}{R1} +\dfrac{sC1}{sC1R2 + 1} +sC2}$$
And,
$$V_O\left[\dfrac{1+sC1R2}{R1} +sC1 +sC2 +s^2C1C2R2\right] = \dfrac{V_I}{R1} + V_O\cdot sC2$$
Therefore,
$$V_O\left[\dfrac{1+sC1R2}{R1} +sC1 +s^2C1C2R2\right] = \dfrac{V_I}{R1} $$
$$V_O\left[1+sC1R2 +sC1R1 +s^2C1C2R1R2\right] = V_I $$
Can you do the final few steps yourself? Do you need help in this any more?