Answer
Maximum:$f(\pm \dfrac{\sqrt 3}{3},\pm \dfrac{\sqrt 3}{3},\pm \dfrac{\sqrt 3}{3})=\dfrac{1}{27}$
Minimum: $f(0,t,\pm \sqrt{1-t^2})=f(t,0,\pm \sqrt{1-t^2})=f(t,\pm \sqrt{1-t^2},0)=0$
Work Step by Step
Use Lagrange Multipliers Method:
$\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$
This yields $\nabla f=\lt \dfrac{2x}{x^2+1},\dfrac{2y}{y^2+1},\dfrac{2z}{z^2+1} \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z\gt$
Using the constraint condition $x^2+y^2+z^2=12$ we get, $\dfrac{2x}{x^2+1}=\lambda 2x, \dfrac{2y}{y^2+1}=\lambda 2y,\dfrac{2z}{z^2+1}=\lambda 2z$
After solving, we get $x=y=z$
Since, $g(x,y)=x^2+y^2+z^2=12$ gives $z=\pm 2$
Thus, $x=y=z=\pm 2$
Hence, Maximum:$f(\pm \dfrac{\sqrt 3}{3},\pm \dfrac{\sqrt 3}{3},\pm \dfrac{\sqrt 3}{3})=\dfrac{1}{27}$
Minimum: $f(0,t,\pm \sqrt{1-t^2})=f(t,0,\pm \sqrt{1-t^2})=f(t,\pm \sqrt{1-t^2},0)=0$
