From the given diagram we can write like
$f=(f_{1}.f_{2})\oplus f_{3}$
We know that $A\oplus B=A\overline{B}+\overline{A}B$
So ,we can expend it and get
$f=(f_{1}.f_{2}).\overline{f_{3}}+\overline{(f_{1}.f_{2})}.f_{3}$
Apply De-Morgan's laws
$\overline{(A.B)}=\overline{A}+\overline{B}$
$\overline{(A+B)}=\overline{A}.\overline{B}$
Now $f=(f_{1}.f_{2}).\overline{f_{3}}+\overline{(f_{1}}+\overline{f_{2})}.f_{3}$
Apply distributive rule
$f=(f_{1}.f_{2}).\overline{f_{3}}+\overline{f_{1}}.f_{3}+\overline{f_{2}}.f_{3}$ -------------$>(1)$
Given that
$f_1=\Sigma(0,2,5,8,14)$
$\overline{f_{1}}=\Sigma(1,3,4,6,7,9,10,11,12,13,15)$
$f_2=\Sigma(2,3,6,8,14,15)$
$\overline{f_{2}}=\Sigma(0,1,4,5,7,9,10,11,12,13)$
$f_3=\Sigma (2,7,11,14)$
$\overline{f_{3}}=\Sigma(0,1,3,4,5,6,8,9,10,12,13,15)$
Now
$f_{1}.f_{2}=\Sigma(0,2,5,8,14).\Sigma(2,3,6,8,14,15)$
Perform Intersection and get it
$f_{1}.f_{2}=\Sigma(2,8,14)$
$(f_{1}.f_{2}).\overline{f_{3}}=\Sigma(2,8,14).\Sigma(0,1,3,4,5,6,8,9,10,12,13,15)$
Perform Intersection and get it
$(f_{1}.f_{2}).\overline{f_{3}}=\Sigma(8)$
and $\overline{f_{1}}.f_{3}=\Sigma(1,3,4,6,7,9,10,11,12,13,15).\Sigma (2,7,11,14)$
Perform Intersection and get it
$\overline{f_{1}}.f_{3}=\Sigma(7,11)$
and last one
$\overline{f_{2}}.f_{3}=\Sigma(0,1,4,5,7,9,10,11,12,13).\Sigma (2,7,11,14)$
Perform Intersection and get it
$\overline{f_{2}}.f_{3}=\Sigma(7,11)$
Now put the values in the equation $(1)$ and get
$f=(f_{1}.f_{2}).\overline{f_{3}}+\overline{f_{1}}.f_{3}+\overline{f_{2}}.f_{3}$
$f=\Sigma(8)+\Sigma(7,11)+\Sigma(7,11)$
Perform Union and get it
$f=\Sigma(7,8,11)$
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This is better way to analyze,how GATE
works.