$\displaystyle f(x,y,z)=3log(x^2+y^2+z^2)-2x^2-2y^3-2z^3$ $(x,y,z)\neq(0,0,0)$ has only one extreme value $\displaystyle log(\frac{3}{e^2})$ Show that the function defined by $\displaystyle f(x,y,z)=3log(x^2+y^2+z^2)-2x^2-2y^3-2z^3$
$(x,y,z)\neq(0,0,0)$ has only one extreme value $\displaystyle log(\frac{3}{e^2})$
Attempt:
$\displaystyle df=\frac{3}{x^2+y^2+z^2}(2xdx+2ydy+2zdz)-4xdx-6y^2dy-6z^2dz$
Then i tried to find the stationary points but couldn't get any further.
 A: Stationary points of $f$ can be found from the system
$$\nabla{f}=\vec{0}$$
which is equivalent to
$$
\begin{cases}
\dfrac{\partial{f}}{\partial{x}}=0, \\
\dfrac{\partial{f}}{\partial{y}}=0, \\
\dfrac{\partial{f}}{\partial{z}}=0, 
\end{cases}
\quad\Leftrightarrow\quad 
\left \lbrace {\matrix{
x\left(\dfrac{3}{x^2 + y^2 + z^2}-2 \right)=0, &(1)\\
y\left(\dfrac{y}{x^2 + y^2 + z^2}-1 \right)=0, &(2)\\
z\left(\dfrac{z}{x^2 + y^2 + z^2}-1 \right)=0. &(3)
 }}\right.
$$
Added:
If $x=0$ and $y=0$ then  $(3)$ implies $z=1.$
Similarly, $x=0,\;\;z=0 \Rightarrow y=1.$
If $y=z=0$  then from $(1)$ we have $x^2=\dfrac{3}{2}\Rightarrow x=\pm{\sqrt{\dfrac{3}{2}}}.$  
The case $x^2 + y^2 + z^2 =\dfrac{3}{2}\;\text{and}\;y=0 $  is impossible because $(3)\Rightarrow z=\dfrac{3}{2},$ but then $ x^2 + y^2 + z^2 \geqslant {\dfrac{9}{4}} >\dfrac{3}{2}. $ Alike the case $x^2 + y^2 + z^2 =\dfrac{3}{2}\;\text{and}\;z=0$  is impossible too.  
One more case is $x=0 \;\text{and}\;\dfrac{y}{x^2 + y^2 + z^2}-1=0 \;\text{and}\;\;\dfrac{z}{x^2 + y^2 + z^2}-1=0$ which leads us to $x=0,\; y=\dfrac{1}{2},\; z=\dfrac{1}{2}.$
Finally, the case $x^2 + y^2 + z^2 =\dfrac{3}{2} \;\text{and}\;\dfrac{y}{x^2 + y^2 + z^2}-1=0 \;\text{and}\;\;\dfrac{z}{x^2 + y^2 + z^2}-1=0$ is also impossible.
