Finding extreme value of function Given $f(x,y,z) = 3 \log( x^2+y^2+z^2) -2x^2-2y^3-2z^3$, $(x,y,z) \neq (0,0,0)$, find the extremum value of $f$.


$f_x = \frac{6x}{x^2+y^2+z^2} - 4x$

$f_y = \frac{6y}{x^2+y^2+z^2} - 6y^2$

$f_z = \frac{6z}{x^2+y^2+z^2} - 6z^2$

How do I proceed ?
Hint: Equating $f_x,f_y,f_z$ to $0$ gives (remember $x^2+y^2+z^2\neq 0$)
$$
\left\{
\begin{aligned}
x(3-2(x^2+y^2+z^2)) &amp;= 0\\
y(1-y(x^2+y^2+z^2)) &amp;= 0\\
z(1-z(x^2+y^2+z^2)) &amp;= 0
\end{aligned}\right.
$$
Now solve by considering cases.
Hint: Equating $f_x,f_y,f_z$ to $0$ gives (remember $x^2+y^2+z^2\neq 0$)
$$
\left\{
\begin{aligned}
x(3-2(x^2+y^2+z^2)) &amp;= 0\\
y(1-y(x^2+y^2+z^2)) &amp;= 0\\
z(1-z(x^2+y^2+z^2)) &amp;= 0
\end{aligned}\right.
$$
Now solve by considering cases.