The output $$Y$$ for the inputs $$A$$ and $$B$$ of circuit is given by
Truth table of the shown circuit is:

$$\text{Output of the first intermediate NAND gate} = \overline{A \cdot B}$$

$$\text{Upper path intermediate NAND gate output} = Y_1 = \overline{A \cdot \overline{A \cdot B}}$$

$$\text{Lower path intermediate NAND gate output} = Y_2 = \overline{B \cdot \overline{A \cdot B}}$$

$$Y = \overline{Y_1 \cdot Y_2} = \overline{\overline{A \cdot \overline{A \cdot B}} \cdot \overline{B \cdot \overline{A \cdot B}}}$$

$$Y = (A \cdot \overline{A \cdot B}) + (B \cdot \overline{A \cdot B})$$

$$Y = (A + B) \cdot \overline{A \cdot B}$$

$$Y = (A + B) \cdot (\bar{A} + \bar{B})$$

$$Y = A\bar{A} + A\bar{B} + B\bar{A} + B\bar{B}$$

$$Y = A\bar{B} + \bar{A}B = A \oplus B \quad (\text{XOR operation})$$