find the points where the tangent line to the function is horizontal y=-x^3+x^2-2

y=-x^3+x^2-2.....the tangent line will be horizontal wherever the derivative = 0

So, the derivative is given by

y ' = -3x^2 + 2x

Set this to 0...so we have.....

-3x^2 + 2x = 0    factor

x(-3x+ 2 ) = 0     set each factor to  0

x = 0   and   x = 2/3

Here's the graph showing the points of interest......https://www.desmos.com/calculator/aewacr9y2d

Thanks to Alan for pointing out that I inadvertantly omitted the leading "-" sign in my previous answer.......I have edited the answer.....(I hope I finally got it correct...!!!)