Given, x2y5=(x+y)7
Taking log on both sides, we get 2logx+5logy=7log(x+y)
On differentiating, we get x2+y5dxdy=x+y7(1+dxdy) ⇒dxdy(x+y7−y5)=x2−x+y7 ⇒dxdy=xy...(i)
Again, differentiating, we get dx2d2y=x2xdxdy−y =x2x⋅(y/x)−y[ { from Eq.(i) } =0