Implicit differentiation question

Find the slope of the tangent line to the curve X^2 + xy + 2y^3 = -254 at the point (4, -5).

Could anyone please tell me how we get to the answer -0.0194805? I'm having difficulty isolating the dy/dx in the equation 2x + y(dy/dx) + 6y^2(dy/dx) = 0.

you need to use the product rule on xy
2x + y + x(dy/dx) + 6y^2(dy/dx) = 0
x(dy/dx) + 6y^2(dy/dx) = -2x - y
dy/dx = -2x -y / (x+6y^2) at (4,-5) = -3/(4+6*25) = -0.0194805

Implicit differentiation provides (don't forget your multiplication rule on the xy term):

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

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

Plug in your point to get:

dy/dx = (-3)/(154) = -.01948...

You are forgetting the product rule in your derivative.

x^2 +xy +2y^3=-254

Derivative: 2x + x(dy/dx) + y + 6y^2(dy/dx) = 0

now, dy/dx= (-y-2x)/(x+6y^2)
Plug in ur points
(5-8)/(154)= -.0194805

x^2 + xy + 2y^3 = -254
2x+x.dy/dx + y +6y^2.dy/dx = 0
At (4,-5)
8 + 4dy/dx -5 +150dy/dx = 0
154dy/dx = -3
dy/dx = -0.0194805

you can use computers do it