what is the equation on the tangent line to the curve 3y + 2xy^2 - 6x = -45 at the point (-2,3)?

Answer 1

3y' + 2y² + 4xyy' - 6 = 0
y'(3+4xy) = 6 - 2y²
y' = (6-2y²) / (3+4xy)
= (6-18) / (3-24)
= 4 / 7

So the equation of the line is:
(y-3) = 4/7(x+2)
y = 4x/7 + 29/7

Answer 2

First use implicit differentiation: 3dy/dx+2y^2dx/dx + 4xydy/dx-6dx/dx=0
solve this for dy/dx=(-2y^2+6)/(4xy+3)
so, when (x,y)=(-2,3), dy/dx=-12/-21=4/7
Finally, just use point slope to get the equation of the tangent line: y-3=4/7(x+2)