Find the gradient of the curve (x^3)-4(y^2)=12xy at the point P(-8,8)

First of all differentiate the equation of the curve implicitly, giving:

3x²-8y(dy/dx)=12y+12x(dy/dx)

=> (dy/dx)(12x+8y)=3x²-12y

=> dy/dx=(3x²-12y)/(12x+8y)

As dy/dx is the gradient of the curve, if we insert x=-8 and y=8, we will have the gradient of the curve specific to the P location:

dy/dx=[3(-8)²-12(8)]/[12(-8)+8(8)]=-3