Answer
$P_{1}(1,1,0,)$ and $P_{2}(-1,-1,0)$
Work Step by Step
The square of the distance from $P(x, y)$ to the origin can be written as
\[
\sqrt{x^{2}+y^{2}+z^{2}}=f(x, y, z)=d^{2}(x, y, z)
\]
Taking into account the comtraint $x y-z^{2}=1(\mathbf{P}$ on the surface), we can use Lagrange Multipliers:
\[
\nabla f=\lambda \nabla g \Rightarrow 2 x \mathbf{i}+2 y \mathbf{j}+2 z \mathbf{k}=\lambda(y \mathbf{i}+x \mathbf{j}-2=\mathbf{k})
\]
So:
\[
\lambda=\frac{2 x}{y}=\frac{2 y}{x}=\frac{2 z}{-2 z}=-1
\]
Using these relationships on the given constraint
\[
\begin{array}{r}
2 x^{2}=2 y^{2} \Rightarrow x=\pm y=-1 \\
x y-z^{2}=1 \Rightarrow \pm x(x)-z^{2}=1 \Rightarrow z=0
\end{array}
\]
Hence, $P_{1}(1,1,0,)$ and $P_{2}(-1,-1,0)$ are points at which d has absolute extrema. They are the points on the surface closest to the origin.
\[
d(\pm 1,\pm 1,0)=\sqrt{1^{2}+1^{2}+0^{2}}=\sqrt{2}=d
\]
