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))#
