$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
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Use $\ds{\tt\mbox{Spherical Coordinates}}$ and
a $\ds{\tt\mbox{Lagrange Multiplier}}$ $\ds{4\mu}$:
\begin{align}
{\cal F}&\equiv r - 4\mu\braces{\bracks{r\sin\pars{\theta}\cos\pars{\phi}}
\bracks{r\sin\pars{\theta}\sin\pars{\phi}}
\bracks{r\cos\pars{\theta}}^{2} -2}
\\[3mm]&=r-4\mu\bracks{r^{4}\sin^{2}\pars{\theta}\cos^{2}\pars{\theta}\sin\pars{\phi}
\cos\pars{\phi} - 2}
\end{align}
$$
{\cal F}=r - \half\,\mu r^{4}\sin^{2}\pars{2\theta}\sin\pars{2\phi} + 8\mu
$$
$$
\begin{array}{rclcrcl}
\partiald{{\cal F}}{r} & = & 0 & \imp &
1 - 2\mu r^{3}\sin^{2}\pars{2\theta}\sin\pars{2\phi} & = & 0
\\[2mm]
\partiald{{\cal F}}{\theta} & = & 0 & \imp &
-\mu r^{4}\sin\pars{4\theta}\sin\pars{2\phi} & = & 0
\\[2mm]
\partiald{{\cal F}}{\phi} & = & 0 & \imp &
-\mu r^{4}\sin^{2}\pars{2\theta}\cos\pars{2\phi} & = & 0
\end{array}
$$
$\ds{\theta \not\in\braces{0,\pi}}$ and $\ds{\phi \not\in\braces{0,\pi,2\pi}}$. That leads to $\ds{\theta =\phi = {\pi \over 4}}$:
$$
2=xyz^{2}={1 \over 8}\,r^{4}\sin^{2}\pars{\pi \over 2}\sin\pars{\pi \over 2}
={r^{4} \over 8}\ \imp\
\begin{array}{|c|}\hline\\
\quad\color{#66f}{\Large r = 2}\quad
\\ \\ \hline
\end{array}
$$