Yes

i suppose the derivative you are talking about is dy/dx, you can find this two ways, explicitly or implicity.

and how do u take the derivative of something like this

3y^2 - 4x=16 ?

implicity:

3y^2 - 4x = 16

=> 6y y' - 4 = 0

=> y' = 4/6y = 2/3y

and if you want, you can solve for y and plug it in

explicitly:

we solve for y first and then take the derivative normally

3y^2 - 4x = 16

=> 3y^2 = 16 + 4x

=> y^2 = (16 + 4x)/3

=> y = sqrt[(16 + 4x)/3] = [(16 + 4x)/3]^(1/2) = (16/3)^(1/2) + (4x/3)^(1/2) .................incorrect, see post below

=> y' = (1/2)(4x/3)^(-1/2) * (4/3) = (2/3)(4x/3)^(-1/2)