Let $ \delta \approx 0^ +$ Let $ \epsilon \approx 0^ +$ Let $ \kappa = \frac {2 - 4\delta - \sqrt {(4\delta - 2 + \epsilon )^2 - 4\cdot 6\delta ^2}}{2}$ We claim that $ \boxed{(x_0,y_0,z_0) = (2\delta + \kappa, - \delta - \delta)}$ satisfies: $ x_0^2 + y_0^2 + z_0^2\approx 0^ +$ and $ f(x_0,y_0,z_0) = 2 - \epsilon\approx 2^ -$ Since $ \delta \approx 0^ +$ and $ \epsilon \approx 0^ +$: $ \kappa = \frac {2 - 4\delta - \sqrt {(4\delta - 2 + \epsilon )^2 - 4\cdot 6\delta ^2}}{2}\approx$ $ \approx \frac {2 - 4\cdot 0 - \sqrt {(4\cdot 0 - 2 + 0)^2 - 4\cdot 6\cdot 0^2}}{2} =$ $ = \frac {2 - \sqrt {2^2}}{2} = 0$ or, $ \kappa\approx 0$ Thus $ x_0 = 2\delta + \kappa\approx 0 ,y_0 = z_0 = - \delta\approx 0^ -$ Hence $ x_0^2,y_0^2,z_0^2\approx 0^ +$ $ \kappa$ is a root of the quadratic $ x^2 + x(4\delta - 2 + \epsilon) + 6\delta ^2 = 0$. Thus $ \kappa$ satisfies $ \kappa ^2 + \kappa (4\delta - 2 + \epsilon) + 6\delta ^2 = 0$ Which is equivalent to: $ \frac {(2\delta + \kappa)^2 + 2\delta ^2}{\kappa} = 2 - \epsilon$ But the LHS is precisely $ f(x_0,y_0,z_0)$ Hence our claim is true.