<s>[INST] $y'(1)$ and $z'(1)$ of $x^2+y^2-2z^2 = 0, x^2+2y^2+z^2 = 4$ Given this equation system 
$$x^2+y^2-2z^2 = 0$$
$$x^2+2y^2+z^2 = 4$$
locally at $(1,1,1)$
It can be solved by:
$$ x^2+y^2-2z^2 = 0  \tag 1$$
$$x^2+2y^2+z^2 = 4 \tag 2$$
$2 \times (1) - (2)$ 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 
$$I(x) = \bigg (x, \sqrt{  \frac {8-3x^2}5} , \sqrt{  \frac {4+x^2}5  } \bigg )    $$
But how can one find out $y'(1)$ and $z'(1)$?
 [/INST] From $x^2-5z(x)^2 = -4$ we derive $2x-10z(x)z'(x)=0.$ With $x=1$ and $z(1)=1$ we get
$$2-10z'(1)=0.$$
Hence $z'(1)=1/5.$
It is now your turn to compute $y'(1)$ with the aid of $3x^2+5y^2 = 8.$
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