Given:
loge(x+y)=4xy
When x=0, then y=1
loge(x+y)=4xy
⇒x+y=e4xy
Now differentiate w.r.t. x
1+y′=e4xy(4y+4xy′)…(i)
At (0,1)⇒y′(0)+1=4⇒y′(0)=3
Now, again differentiate equation (i), we get
y"=e4xy(4y+4xy′)2+e4xy(4y′+4y′+4xy")
At (0,1)
y"(0)=1(4×1+0)2+1(4×3+4×3+0)
⇒y"(0)=16+24=40
⇒y"(0)=40