calculus
posted by jump .
The slope of the curve x^3y^2 + 2x  5y + 2 = 0 at the point (1,1) is

using implicit derivatives :
x^3(2y)dy/dx + y^2 (3x^2) + 2  5 dy/dx = 0
dy/dx(2x^3 y  5) = 3x^2 y^2  2
dy/dx = (3x^2y^2  2)/(2x^3y  5)
at the point (1,1)
dy/dx = (3  2)/(25) = 5/3
now you have the slope of the tangent, and a point on that tangent.
Use the method you commonly use to find the equation of that straight line