We know the tangent line is horizontal when y'=0. So we want to find all points on the curve where y'=0.
STEP 1: Use implicit differentiation to find y'
2x + (1*y + xy') + 2y*y' = 0 = 2x + y + xy' + 2yy' = 0
STEP 2: We are looking for where y'=0, so go ahead and plug 0 in for y' in the equation above.
2x + y + x(0) + 2y(0) = 0
y = -2x
STEP 3: Now we know that we have a horizontal tangent line whenever y=-2x. But the question is asking us "for what points." To find the points, we are looking for the points on the curve for which y=-2x.
STEP 4: When does y=-2x on the curve x^2 + xy + y^2 = 1?
To solve this question, we can sub in -2x wherever we see a y in our original equation (substitution method).
x^2 + x(-2x) + (-2x)^2 = 1
x^2 -2x^2 +4x^2 = 1
3x^2 = 1
x^2 = 1/3
x = +- sqrt(1/3)
STEP 5: Now that we know the x-value of the point, we can easily find the y-value of the point because we know y=-2x where the tangent line is horizontal.
POINTS: (sqrt(1/3),-2sqrt(1/3)) and (-sqrt(1/3),2sqrt(1/3))