(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).