(a) Assume that #2/x+2/y=7#, or #2x^{-1}+2y^{-1}=7#, implicitly defines #y# as a function of #x# and differentiate both sides of this equation with respect to #x# to get #-2x^{-2}-2y^{-2}*y'=0# (the Chain Rule is used here in the second term).
This can be rearranged to say that #-(y')/(y^2)=1/(x^2)#, or #y'=-(y^2)/(x^2)#.
(b) Do algebra to solve #2/x+2/y=7# as follows: #2/y=7-2/x=(7x-2)/x ==> y/2=x/(7x-2) ==> y=(2x)/(7x-2)#.
Now differentiate with the Quotient Rule:
#y'=((7x-2)*2-(2x)*7)/((7x-2)^2)#
#=(14x-4-14x)/((7x-2)^2)=-(4)/((7x-2)^2)#
(c) Finally, substitute #y=(2x)/(7x-2)# into the equation #y'=-(y^2)/(x^2)# and simplify:
#y'=-(((2x)/(7x-2))^2)/(x^2)=-(4x^2)/((7x-2)^2)*1/(x^2)=-(4)/((7x-2)^2)#, which is the same as the answer from part (b).
