Answer
Maximum:$f(2,2\sqrt2,2 \sqrt2)=e^{16}$, Minimum: $f(-2,-2\sqrt2,-2 \sqrt2)=e^{-16}$
or,
Maximum:$f(2,\sqrt8, \sqrt 8)=e^{16}$, Minimum: $f(-2,-\sqrt 8,-\sqrt 8)=e^{-16}$
Work Step by Step
Use Lagrange Multipliers Method:
$\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$
This yields $\nabla f=\lt yze^{xyz},xze^{xyz},xye^{xyz} \gt$ and $\lambda \nabla g=\lambda \lt 4x,2y,2z\gt$
Using the constraint condition $2x^2+y^2+z^2=24$ we get, $yze^{xyz}=\lambda 4x, xze^{xyz}=\lambda 2y,xye^{xyz}=\lambda 2z$
After solving, we get $y=z=2x^2$
Since, $g(x,y)=2x^2+y^2+z^2=24$ gives $x=\pm 2$
Thus, $y=z=\pm 2\sqrt2$
Hence, Maximum:$f(2,2\sqrt2,2 \sqrt2)=e^{16}$, Minimum: $f(-2,-2\sqrt2,-2 \sqrt2)=e^{-16}$
or,
Maximum:$f(2,\sqrt8, \sqrt 8)=e^{16}$, Minimum: $f(-2,-\sqrt 8,-\sqrt 8)=e^{-16}$
