Calculus, implicit diff help.........

can anyone answer these questions with step by step method? thanks alot!!!!!!

1. Find d^2y/dx^2,
y^4 + x + sin(y) = 7

dy/dx =1/(-4*y^3-cos(y))

i tried several ways and didn't get it right, these answers i tried and wrong..
d^2y/dx^2 = (12y^2-sin(y))/(4y^3+cos(y))^2
d^2y/dx^2 = -(12y^2-sin(y))/(4y^3+cos(y))^2
d^2y/dx^2 = -(sin(y)-12y^2)/(4y^3+cos(y))^2
and what is the answer if these are not correct?

2. Given the function
y = ( x+4 ) ( x^2−5 x )
find the coordinates of the two stationary points and the point of inflection.
Note. A stationary point is a critical point at which the derivative is 0.
answer should have the form:
[ p(a, b), p(c, d), p(e, f) ]
for some constants a, b, c, d, e, f, which should be given to one decimal place accuracy (recall the square brackets make it a list of pairs). The final pair should be the inflection point.

my last answer was [ p(2.9, -480.4), p(-2.3, 665.8), p(0.3, 13.6) ]

Your answer is partially correct.
For your 1st point, the 2nd coordinate is incorrect.
For your 2nd point, the 2nd coordinate is incorrect.
For your 3rd point, the 2nd coordinate is incorrect.

Hint: the stationary points occur for points on the curve where the first derivative is zero.

1)

y^4 + x + sin(y) = 7
d/dx [y^4 + x + sin(y)] = d/dx (7)
d/dx y^4 + d/dx x + d/dx sin(y) = 0
(dy/dx) d/dy y^4 + 1 + (dy/dx) d/dy sin(y) = 0
(dy/dx) (4y^3) + 1 + (dy/dx) cos(y) = 0
(dy/dx) (4y^3 + cos(y)) = - 1
dy/dx = -(4y^3 + cos(y))^(-1)
d^2 y/dx^2 = d/dx -(4y^3 + cos(y))^(-1)
d^2 y / dx^2 = (4y^3 + cos(y))^(-2) * (dy/dx) * (d/dy (4y^3 + cos(y)))
d^2 y / dx^2 = (4y^3 + cos(y))^(-2) * (-(4y^3 + cos(y))^(-1)) * (12y^2 - sin(y))
d^2 y / dx^2 = -(4y^3 + cos(y))^(-3) * (12y^2 - sin(y))
d^2 y / dx^2 = -(12y^2 - sin(y)) / (4y^3 + cos(y))^(3)
d^2 y / dx^2 = (sin(y) - 12y^2) / (4y^3 + cos(y))^(3)

2)