Finding second derivatives?

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

differentiate both sides with respect to x

2(3x^2) - 2(y+x dy/dx) + 4(2y) dy/dx = 0

6x^2-2y-2x dy/dx +8y dy/dx = 0

6x^2-2y+dy/dx(8y-2x)=0

dy/dx(8y-2x)= -6x^2+2y

dy/dx = (2y-6x^2) / (8y-2x)

Apply the quotient rule to find d^2y/dx^2

d^2y/dx^2 = [ (8y-2x) (2 dy/dx - 12x) - (2y-6x^2)(8 dy/dx - 2x)] / (8y-2x)^2

d^2y/dx^2 = [ (16y dy/dx -96xy-4xdy/dx+24x^2) - (16ydy/dx-4xy-48x^2dy/dx+12x^3)] / (8y-2x)^2

d^2y/dx^2 = [ 16y dy/dx-96xy-4xdy/dx+24x^2-16ydy/dx+4xy+48x^2dy/dx-12x^3)/(8y-2x)^2

d^2y/dx^2 = [ dy/dx(-4x+48x^2)-92xy+24x^2-12x^3)] /(8y-2x)^2

plug dy/dx = (2y-6x^2) / (8y-2x) into the above equation

d^2y/dx^2 = [ (2y-6x^2){(48x^2-4x) / (8y-2x)} -92xy+24x^2-12x^3] /(8y-2x)^2

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Is your first question printed incorrectly?

Is it 4x^2+xy-3y=5 ?

Find the first derivative, then take the derivative of that. That is how the second (and third and so on) derivative is defined.