The curve is defined by `x^3*y^2 = cos(pi*y)`

Substituting x = -1 and y = 1 gives -1 = -1, proving that (-1, 1) lies on the curve `x^3*y^2 = cos(pi*y)`

The slope of the tangent to the curve at the point (-1, 1) is given by `dy/dx` at (-1,...

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The curve is defined by `x^3*y^2 = cos(pi*y)`

Substituting x = -1 and y = 1 gives -1 = -1, proving that (-1, 1) lies on the curve `x^3*y^2 = cos(pi*y)`

The slope of the tangent to the curve at the point (-1, 1) is given by `dy/dx` at (-1, 1)

Using implicit differentiation

`3*x^2*y^2 + x^3*2y*(dy/dx) = sin(pi*y)*pi*(dy/dx)`

=> `(dy/dx) (2*x^3*y - sin(pi*y)*pi) = -3*x^2*y^2`

=> `dy/dx = (3*x^2*y^2)/(sin(pi*y)*pi - 2*x^3*y)`

`dy/dx` at (-1, 1) is

`(3*1*1)/(sin pi*pi + 2)`

=> `3/2`

**The slope of the tangent to the curve `x^3*y^2 = cos(pi*y)` at (-1, 1) is **`3/2`