Answer
$f$ has one critical point at $P = \left( {0,0} \right)$.
Since ${f_x}$ and ${f_y}$ do not exist at $P = \left( {0,0} \right)$, $f$ is nondifferentiable at $P$.
$f$ has a minimum value at $P$.
Work Step by Step
We have $f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} $.
Since the right hand side is non-negative, the minimum value of $f$ is $0$. Thus, the critical point is $P = \left( {0,0} \right)$. We conclude that $f$ has a minimum value at $P$.
The partial derivatives are
${f_x} = \frac{x}{{\sqrt {{x^2} + {y^2}} }}$, ${\ \ }$ ${f_y} = \frac{y}{{\sqrt {{x^2} + {y^2}} }}$
However, ${f_x}$ and ${f_y}$ do not exist at $P = \left( {0,0} \right)$. Therefore, $f$ is nondifferentiable at $P$.
