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## Homework Statement

Find the points on the graph of x

^{3}-y

^{3}=3xy-3 where the tangent line is horizontal

## Homework Equations

y = f(x) so implicit differentiation must be used when taking the derivative of y

(xy)' = xy' + y

## The Attempt at a Solution

So if the tangent line is horizontal the slope = 0 and the derivative of y' must be equal to zero.

So first taking the derivative I end up with

3x

^{2}-3y

^{2}y' = 3xy' + 3y

Factor out the three and divide by three on both sides to cancel them out. That leaves me with

x

^{2}-y

^{2}y' = xy' + y

Subtract y on both sides, add y

^{2}y' on both sides and get

x

^{2}-y = xy' + y

^{2}y'

Factor out the y' to get

x

^{2}-y = y'(x+y

^{2})

divide by x+y on both sides and

y' = (x

^{2}-y)/(x+y

^{2})

So at this point, I set y' equal to zero and solve for x? Don't I remove the y variables or something like that?

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