Question

find the first and second derivatives for 3x^4-4xy^2+7=2y

Answer 1

3x^4-4xy^2+7=2y

12x^3 - 4y^2 - 8xyy' = 2y'
(8xy+2)y' = 12x^3-4y^2
y' = (6x^3-3y^2)/(4xy+1)

y" =
(18x^2-6yy')(4xy+1) - (6x^3-3y^2)(4y+4xy')
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(4xy+1)

Now it's just algebra. Substitute the value of y' in the above expression and then massage it as you will.

Answer 2

oops. The denominator is (4xy+1)^2

Answer 3

See if you can get what wolframalpha.com gets:

http://www.wolframalpha.com/input/?i=%28%2818x^2-6y%28%286x^3-3y^2%29%2F%284xy%2B1%29%29%29%284xy%2B1%29+-+%286x^3-3y^2%29%284y%2B4x%28%286x^3-3y^2%29%2F%284xy%2B1%29%29%29+%29%2F%284xy%2B1%29+^2