Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)/Explicit Method

Proof
First we make sure that $\tuple {1, -2, -2}$ actually satisfies the equation:

and it is seen that this is the case.

When $z = -2$ we have that $\sqrt {x^2 - 2 x y - 1} < 0$.

Hence:
 * $z = -\sqrt {x^2 - 2 x y - 1}$

Thus we continue: