Finding the slope of tangent line at given point?

Question

I need to find the the slope of the line tangent to 3x^2+xy-4y^3=108 which also passes through point (6,0).

Answer 1

Slope m = dy/dx

Given line equation  3x^2+xy-4y^3 = 108

Apply derivative to each side.

6x+[xdy/dx+y]-12y^2dy/dx = 0

dy/dx[x-12y^2] = -6x-y

dy/dx = (-6x-y)/(x-12y^2)

At the point x = 6, y = 0.

m = (-36-0)/(6-0)

m = -36/6

m = -6

Slope of the given line is -6.