If x^2 + 2xy + 3y^2 = 1, show that y^2 = -2/(x + 3y)^3

Question

If x² + 2xy + 3y² = 1, show that y² = \(\frac{-2}{(x + 3y)^3}\)

Answer 1

x² + 2xy + 3y² = 1 

Differentiate w.r.t x we get 

2x + 2(x.y₁ + y.1) + 6y.y₁ = 0 

y₁(2x + 6y) = -2x - 2y