For the same input, exor gate will produce $0$ whereas for different inputs it will produce $1$ as output. Here is $f_1,f_2$ $2,4$ as common minterm so it will not be in $f_1\oplus f_2$. all the minterm not in common will be in $f_1\oplus f_2$.(Either or)

so $f_1\oplus f_2=\sum(0,1,3,6)$.

Method 2)

Since we know that AND gate takes common minterm, OR gate takes all minterm, and NOT gates take those minterms that are not present.

$f_1\oplus f_2=\bar f_1f_2+f_1\bar f_2$

$\bar f_1=(0,5,6,7),f_2=(0,2,4,6)\rightarrow \bar f_1f_2=(0,6)$ (As AND gate takes common minterm)