The slope of the tangent to the curve y^3x + y^2x^2 = 6 at (2,1)

Y^3x + y^2x^2 = 6

Using implicit differentiation

3 y^2. y'. x+y^3+2 y. y'. x^2+2 y^2. x=0

y' (3 y^2. x+2 y. x^2) = - y^3-2 y^2. x

y' = - [y^2 (y+2 x) ]/[x y (3 y+2 x) ]

= - [y (y+2 x) ]/[x (3 y+2 x) ]

Substituting x = 2 and y = 1:

y' = - (1) (1+4) / [2 (3+4) ]

= - 5/14

The slope of the tangent of the curve at (2,1) will be:

m = - 5/14