Find the point on the curve $x^{2} - yz=1$ that is closest to the origin The problem statement is as follows:
Find the point on the curve $x^{2} - yz=1$ that is closest to the origin.
To do this we minimize $D^{2} = x^{2} + y^{2} + z^{2}$ with the above constraint.
Attempted solution:
Isolate $z$ in the constraint:
$$
x^{2} - yz = 1 \implies z = \frac{x^{2}-1}{y}
$$
Substitute into distance formula:
$$
x^{2} + y^{2} + \left(\frac{x^{2}-1}{y}\right)^{2}
$$
Differentiate in terms of x and y and set to zero
$$
\frac{\partial D^{2}}{\partial x} = 2x+4x\left(\frac{x^{2}-1}{y^{2}}\right)=0,\frac{\partial D^{2}}{\partial y} = 2y-2\left(\frac{(x^{2}-1)^{2}}{y^{3}}\right)=0
$$
which has real solutions $x=0, y=+1$, giving the critical point as $x=0, y=1, z=-1$.
The correct answer is $x=\pm 1, y=0, z=0$.
What am I doing wrong?

An alternative approach is using Lagrange multipliers. Construct the Lagrangian auxiliary function
$$\mathcal L=x^{2} + y^{2} + z^{2}+\lambda\left[1-\left(x^{2} - yz\right)\right]$$
and solve the first order conditions
$$\begin{aligned}
\frac{d\mathcal L}{d x}=2x-2x\lambda=0\\
\frac{d\mathcal L}{d y}=2y+z\lambda=0\\
\frac{d\mathcal L}{d z}=2z+y\lambda=0\\
\frac{d\mathcal L}{d \lambda}=1-x^2+yz=0\\
\end{aligned}$$
that, of course, result in $x=\pm 1$, $y=0$ and $z=0$.
