Given,
(1+e2x)dxdy+2(1+y2)ex=0
⇒1+y2dy+1+e2x2exdx=0 .........(i)
Now integrating both side we get,
⇒∫1+y2dy+∫1+e2x2exdx=∫0
⇒tan−1y+2tan−1ex=c
∵y(0)=0
so, C=2π⇒tan−1y+2tan−1ex=2π ....(ii)
Now from equation (i), we get (dxdy)x=0=−1
From equation (ii) we get, y (In 3)=−31
So, 6[y′(0)+(y In 3)2]=6[−1+31]=−4.