Answer
Minimum: $f(\pm 1, \pm 1)=2$; No maximum value.
Work Step by Step
Use Lagrange Multipliers Method:
$\nabla f(x,y)=\lambda \nabla g(x,y)$
This yields $\nabla f=\lt 2x,2y \gt$ and $\lambda \nabla g=\lambda \lt y,x \gt$
Using the constraint condition $xy=1$ we get, $2x=\lambda y, 2y=\lambda x$
After solving, we get $x=y=1$ and $\lambda =2$
Since, $g(x,y)=xy=1$ yields $x=y=\pm 1$
Hence, Minimum: $f(\pm 1, \pm 1)=2$; No maximum value.