How can one show that this equation system
$$x^2+y^2-2z^2 = 0$$
$$x^2+2y^2+z^2 = 4$$
locally at $(1,1,1)$ can be solved through the differentiable functions $y(x)$ and $z(x)$?
I put it in WolframAlpha, but I get weird results and I don't know what the "algorithm" is to solve it through differentiable functions...
$$ x^2+y^2-2z^2 = 0 \tag 1$$
$$x^2+2y^2+z^2 = 4 \tag 2$$
$2 \times (1) - (2)$ gives $$x^2-5z^2 = -4 $$
$2 \times (2) + (1)$ gives $$3x^2+5y^2 = 8 $$
In the neighbourhood of $(1,1,1)$ the intersection can be parametrized as
$$I(x) = \bigg (x, \sqrt{ \frac {8-3x^2}5} , \sqrt{ \frac {4+x^2}5 } \bigg ) $$