Answer
The critical points are points in the $y$-axis. Hence, there are infinitely many critical points.
The minimum value of $f$ is $0$.
$f$ does not have any local maxima.
Work Step by Step
We have a function of two variables: $f\left( {x,y} \right) = {x^2}$.
The partial derivatives are
${f_x} = 2x$, ${\ \ \ }$ ${f_y} = 0$
${f_{xx}} = 2$, ${\ \ }$ ${f_{yy}} = 0$, ${\ \ }$ ${f_{xy}} = 0$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 2x = 0$, ${\ \ \ }$ ${f_y} = 0$
The solution is $x=0$ and $y \in \mathbb{R}$. Thus, the critical points are points in the $y$-axis. Hence, there are infinitely many critical points.
The discriminant $D$ is $D = {f_{xx}}{f_{yy}} - {f_{xy}}^2 = 0$. Therefore, by Theorem 2, the Second Derivative Test fails for all of them.
Since ${x^2} \ge 0$, the minimum value of $f$ is $0$.
Since $D=0$ and ${f_{xx}} > 0$, by Theorem 2, $f$ does not have any local maxima.
