(x2+y2)2 = xy ... (i)
Differentiating with respect to x, we have,
2 (x2+y2)(2x+2y.

dy

dx

) = y +

xdy

dx

⇒ 4x (x2+y2) + 4y (x2+y2).

dy

dx

= y +

xdy

dx

⇒

dy

dx

(4x2y+4y3−x) = y - 4x3−4xy2
⇒

dy

dx

=

y−4x3−4xy2

4x2y+4y3−x

y 3cos(logx) 4sin(logx)
Differentiating the above function with respect to x, we have,

dy

dx

=

−3sin(logx)

x

+

4cos(logx)

x

x

dy

dx

= - 3 sin (logx) + 4 cos (log x)
Again differentiating with respect to x, we have,
x

d2y

dx2

+

dy

dx

=

−3cos(logx)

x

-

4sin(logx)

x

⇒ x2

d2y

dx2

+

dy

dx

+ x

dy

dx

= - (3 cos (log x) + 4 sin (log x))
⇒ x2

d2y

dx2

+

dy

dx

+ x

dy

dx

= - y
⇒ x2

d2y

dx2

+

dy

dx

+ x

dy

dx

+ y = 0