Implicit differentiation is a special case of the chain rule for derivatives. Generally differentiation problems involve functions i.e. #y=f(x)# - written explicitly as functions of #x#.

However, some functions y are written implicitly as functions of #x#. So what we do is to treat #y# as #y=y(x)# and use chain rule. This means differentiating #y# w.r.t. #y#, but as we have to derive w.r.t. #x#, as per chain rule, we multiply it by #(dy)/(dx)#.

In the given instance, we have #x^2=(4x^2y^3+1)^2#. Now differentiating both sides **w.r.t.** #x# and remembering to multiply by #(dy)/(dx)#, whenever we differentiate w.r.t. #y#, we get

#2x=2(4x^2y^3+1)xx(4xx(2x xx2y^3+x^2xx3y^2xx(dy)/(dx)))#

or #2x=(16xy^3+12x^2y^2(dy)/(dx))# or

#(16xy^3+12x^2y^2(dy)/(dx))=(2x)/(2(4x^2y^3+1))=x/(4x^2y^3+1)# or

#12x^2y^2(dy)/(dx)=x/(4x^2y^3+1)-16xy^3# or

#(dy)/(dx)=x/(12x^2y^2(4x^2y^3+1))-(16xy^3)/(12x^2y^2)#

= #1/(12xy^2(4x^2y^3+1))-(4y)/(3x)#