When the derivative of the function in a point is , its tangent is horizontal.

Your function is

Take the derivative of this function (using the sum and power differentiation principles) :

Now you are looking for x-coordinates when the tangent is horizontal. Thus, you want to solve :

It factorizes ! , :

Thus, has an horizontal tangent in x-coordinates and . But you also want y-coordinates, right ? So plug this back into your first equation :

Thus, the curve has horizontal tangents in coordinates , and , that is, and .

So you are right (prefer keeping fractional answer though).