If x^2=25-y^2, what is the value of [(d^2)(y)]/(dx^2) at the point (3,4)?

2x = -2y dy/dx

dy/dx = -x/y
at (3,4)
dy/dx = -3/4

differentiate again,
2 = -2y (d^2 y/dx^2) + (dy/dx)(-2dy/dx)
2 = -2y (d^2 y/dx^2) - 2(dy/dx)^2
2 = -2(4)(d^2 y/dx^2) - 2(9/16)

(d^2 y/dx^2) = (2 + 9/8)/-8 = -25/64

or

y^2 = 25-x^2
y = (25-x^2)^(1/2)
dy/dx = (1/2)(25-x^2)^(-1/2) (-2x)
= -x(25-x^2)^(-1/2)
d^2 y/dx^2 = -x(-1/2)(25-x^2)^(-3/2) (-2x) - (25-x^2)^(-1/2)
= -x^2/(25 - x^2)^(3/2) - 1/(25-x^2)^(1/2)
= -9/16^(3/2) - 1/16^(1/2)
= -9/64 - 1/4
= -25/64