operational amplifier - Is this correct for non-inverting op-amp?

Do 0.1 uf and 100 k ohm act as low-pass filter in here?

Answer

Do 0.1 uf and 100 k ohm act as low-pass filter in here?

No, since there is no current through the parallel RC network, there is zero volts across the network and so

$$V_{out} = V_- = V_+ = V_{in}$$

thus, the circuit implements a unity gain buffer:

$$V_{out} = V_{in}$$

Another way to see this is to consider the canonical non-inverting configuration in the frequency domain:

The transfer function (frequency response) is:

$$\dfrac{V_{out}}{V_{in}} = 1 + \dfrac{Z_2}{Z_1}$$

In the schematic you provide, $Z_1$ is an open circuit, i.e. $Z_1 = \infty $. Thus, the transfer function for your circuit is:

$$\dfrac{V_{out}}{V_{in}} = 1 + \dfrac{Z_2}{\infty} = 1$$

which is, as before, just a unity gain buffer. Now, if you place a resistor $R_1$ to ground from the inverting input, the transfer function is:

$$\dfrac{V_{out}}{V_{in}} = 1 + \dfrac{100k\Omega || \frac{1}{j \omega (0.1\mu F)}}{R_1}$$

which will provide some filtering - the details of which I will leave as an exercise for the reader.

:)