`6x^2+3y^2=8`

`3y^2 = 8-6x^2`

`y^2 = (8-6x^2)/3`

Let us derivate both sides with respect to x;

`2y(dy)/dx = 1/3(0-6xx2x)`

`(dy)/dx = -4x/(2y)`

`(dy)/dx = (-2x)/y`

`2y(dy)/dx = 1/3(0-6xx2x)`

`2y(dy)/dx = -4x`

`y(dy)/dx = -2x`

Derivate both sides with respect to x;

`yxx(d^2y)/(dx^2)+(dy)/dx*(dy)/dx = -2`

`(d^2y)/(dx^2) = (-2-((dy)/dx)^2)/(y)`

`(d^2y)/(dx^2) = (-2-(4x^2)/y^2)/y`

`(d^2y)/(dx^2) = (-2(y^2+2x^2))/y^3`

*So the answers are;*

`(dy)/dx = (-2x)/y`

`(d^2y)/(dx^2) = (-2(y^2+2x^2))/y^3`

`6x^2 + 3y^2=8`

`` Take a derivative of this equation with respect to x. Remember to use the chain rule to find the derivative of y squared.

`12x + 6y(dy/dx) = 0`

Express dy/dx from this equation: `dy/dx = -2x/y`

Take the derivative again to obtain the second derivative. Use the quotient rule for the x/y fraction:

`d^2y/dx^2 = -2(y - x(dy/dx))/y^2`

`` You can now substitute the previous expression for dy/dx in here to get the expression for the second derivative in terms of x and y:

`(d^2y/dx^2)=-2(y+2x^2/y)/y^2=-2(y^2+2x^2)/y^3`

``

``