Calculus: Early Transcendentals, Chapter 3, 3.5, Section 3.5, Problem 35

Note:- 1) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = real number
2) If y = u*v ; where both u & v are functions of 'x' , then
dy/dx = u*(dv/dx) + v*(du/dx)
3) If y = k ; where 'k' = constant ; then dy/dx = 0
Now, the given function is :-
9(x^2) + (y^2) = 9
Differentiating both sides w.r.t 'x' we get;
18x + 2y(dy/dx) = 0
or, 9x + y(dy/dx) = 0 .........(1)
or, dy/dx = -(9x)/(y)..........(2)
Differentiating (1) again w.r.t 'x' we get
9 + {(dy/dx)^2} + [(y)*y"] = 0..........(3)
Putting the value of dy/dx from (2) in (3) we get
9 + 81{(x^2)/(y^2)} + [(y)*y"] = 0
or, y" = -[(9{1 + 9(x^2)/(y^2)}/(y)
or, y" = -[9{(y^2) + 9(x^4)}]/(y^3)