Equating like coefficients yields
y = 12λx and x = 2λy.
==> y = 12λ(2λy) = 24λ^2y.
(i) If y = 0, then 6x^2 + 0 - 8 = 0 ==> x = ± 2/√3.
(ii) Otherwise, 1 = 24λ^2 ==> λ = ± 1/√24.
This yields y = ±12x/√24 = ± x√6.
Plug this into g: 6x^2 + 6x^2 - 8 = 0 ==> x = ±√(2/3).
This yields (x, y) = (±√(2/3), ±2), (±√(2/3), ∓2).
Testing the critical points:
f(± 2/√3, 0) = 0
f(±√(2/3), ±2) = 2√(2/3) <----Maximum
f(±√(2/3), ∓2) = -2√(2/3) <----Minimum