Question

Find all points on the curve `x^2y^2+xy=2` where the slope of the tangent line is `-1`?

Answer 1

Use Implicit Differentiation (and hope it all falls out the bottom)

Differentiate both sides of the equation:
`D (x^2y^2 + xy) = D(2)`

using the product rule (a couple of times) and the chain rule:
`(2x*y^2 +x^2*2y*y') + ((1)*y + xy') = 0`

giving
`y' (2yx^2 + x) + 2xy^2 + y = 0`
or
`y' = - (2xy^2 + y)/(2yx^2 + x)`

We are asked for the points where the slope (`y'`) equals `- 1`
so
`- (2xy^2 + y)/(2yx^2 + x) = - 1`
or
`2xy^2 + y = 2yx^2 + x`

`y (2xy + 1) = x (2xy + 1)`

provided `xy != - 1/2` (which can be ruled out as extraneous by referring to the original equation
we have
`y = x`

The original equation becomes
`x^2*x^2 + x*x = 2`

`x^4 + x^2 - 2 = 0`

`(x^2+2)*(x^2-1) = 0`

since `x^2+2 = 0` is obviously extraneous
`x^2 = +- 1`

with (as already determined `y=x`)
the required points are
`(-1,-1)` and `(1,1)`