math calculus
posted by LEON .
Use implicit differentiation to find the equation of the tangent line to the curve xy^3+2xy=9 at the point (31). The equation of this tangent line can be written in the form y=mx+b where m is

x(3y^2dy) + y^3(dx) +2xdy + 2ydx = 0
dy (3xy^2+2x) = dx (y^3+2y)
dy/dx = (y^3+2y)/(3xy^2+2x)
I assume you mean the point (3,1)
x = 3
y = 1
so
m = dy/dx = (1+2)/(9+6) =  3/15 = 1/5
put in point
1 = m * 3 + b
1 = 3/5 + b
b = 8/5
y = x/5 + 8/5
5y = 8x
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