Lagrange multipliers work just fine here. But the Arithmetic Mean/Geometric Mean Inequality is a useful alternative that is calculus-free. Let $a=x^2$, $b=y^2$, and $c=z^2$. Then by AM/GM $\frac{a+b+c}{3}\ge \sqrt[3]{abc}.$ Conveniently, in this case the inequality will give you both the maximum and the minimum.
Remark: To use Lagrange multipliers, first note that the max and min are positive and negative respectively. So each occurs on the boundary. (For if not, scale until we reach the boundary.) The usual partial derivative calculation leads to the equations $yz+2\lambda x=0$, $xz+2\lambda y=0$, $xy+2\lambda z=0$, and $x^2+y^2+z^2=3$.
Exploit the symmetry by multiplying the first equation through by $x$, the second by $y$, the third by $z$. We conclude that $2\lambda x^2=2\lambda y^2=2\lambda z^2$, and now things become simple.
If $\lambda=0$, then two of $x$, $y$, $z$ must be $0$, and the third must be $\pm\sqrt{3}$ (but these are irrelevant for the max or min). If $\lambda\ne 0$, then $x^2=y^2=z^2=1$.
Another way to look at the problem, fairly natural but somewhat less attractive than AM/GM or Lagrange multipliers, is to use spherical coordinates. Note that the general point at distance $r$ from the origin has coordinates $\,x=r\sin\theta\cos\varphi$, $\,y=r\sin\theta\sin\varphi$, $\,z=r\cos\theta$. Calculate the product $xyz$. The max/min problem now decomposes into two one-variable problems.
All of our arguments have been computational. Note that a non-computational approach that gives the answer in seconds has been posted.