Find where the tangent is parallel to the x axis on this curve?

If the tangent line is parallel to the x-axis, its slope is zero.

If the tangent line is parallel to the y-axis, its slope is undefined (which here means where division by zero would occur)

And as we all know, the slope of the tangent line at some point P equals the value of the derivative at P (except for the dreaded "division by zero" case).

Using implicit differentiation,

x² + xy + y² =7

2x + y + xy' + 2yy' = 0

Solving for y',

y' = -(2x + y) / (x + 2y)

So if y' = 0, then 2x + y = 0, or y = -2x. Substituting this into the equation of the curve,

x² + x(-2x) + (-2x)² =7

3x² = 7

x = ±√(7/3) = ±√21 / 3

Then, since y = -2x, the points where the tangent line has slope 0 (and so is parallel to the x-axis) are (√21 / 3, -2√21 / 3) and (-√21 / 3, 2√21 / 3)

For (b), start by setting the denominator of y' to 0, and substitute like I did. Because of symmetry, I can tell without solving that you should get (2√21 / 3, -√21 / 3) and (-2√21 / 3, √21 / 3).

a) ok when the tangent is parallel to the x-axis the line in horizontal and the slope is zero therefore the derivative must be equal to zero

ok so take the derivative of the equation to get get

2x + x(dy/dx) + y + 2y(dy/dx) = 0

dy/dx = (-2x-y)/(x+2y)

set dy/dx = 0 (slope = 0)

0 = (=2x-y)/(x+2y)

solve for x and y

0 = -2x - y

using a system of equations related the above to x^2 + xy + y^2 = 7

we get x = (7/3)^(1/2) and y = -2(7/3)^(1/2)

b) when the tangent is parallel to the y-axis that means the slope is undefined and therefore the bottom of our equaion is equal to zero so

0 = x+2y using a system of equations related again to x^2 + xy + y^2 = 7

we get...

y = (7/3)^(1/2) and x = -2(7/3)^(1/2)

Hope I helped!