Answer
Maximum:$f(2, 2,2)=f(-2,-2,-2)=3 \ln (5)$
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:$f(2, 2,2)=f(-2,-2,-2)=3 \ln (5)$
