how do you solve this calculus question?

Question

find dy/dx

x^2 + xy + y^2 = 7

Answer 1

x^2 + xy + y^2 = 7
dx^2/dx + dxy/dx + dy^2/dx = 0
2x + y + xdy/dx + 2ydy/dx = 0
xdy/dx + 2dy/dx = -2x - y
(x + 2)dy/dx = -2x - y
dy/dx = (-2x - y)/(x + 2y)

Answer 2

differetiate:
2xdx+xdy+ydx+2ydy=0
put the dx stuff on one side and the dy stuff on the other side of the equals sign. Factor out so you have Ady=Bdx
Then dy/dx=B/A

Answer 3

2x + xdy/dx +y +2ydy/dx = 0
solve for it now

Answer 4

differentiate the equation by dx, you get (remember dx/dx = 1)

2x(dx/dx) +y(dx/dx) +x(dy/dx) + 2y(dy/dx) = 0, then by grouping

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

dy/dx = - (2x+y)/(x+2y) which is likely an ellipse offset from 0,0

Answer 5

x^2 + xy + y^2 = 7

differentiating implicitly

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

y'(x + 2y) + 2x + y = 0

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

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