#### Answer

Maximum value is $1$
Minimum value is $-1$

#### Work Step by Step

$f(x,y,z)=xyz$; $x^2+y^2+z^2=3$
According to Lagrange multipliers, we have the following equations:
$yz= \lambda (2x)$ ...(1)
$xz= \lambda (2y)$... (2)
Divide equation (1) by (2), we get
$y^2=x^2$ and $x^2=z^2$
Thus, $x^2=y^2=z^2$
Therefore,$x^2+x^2+x^2=3$ or $x^2=1$
Maximum value is $1$
Minimum value is $-1$