Answer: D

Solution:

Put $t=|x|.$ The second equation gives $t+|y|=3$ or $5,$ while the first gives $y=3-\frac{t^2}{3}.$

If $y\ge0,$ then $0\le t\le3.$ For $t+y=3,$ substitution gives $t-\frac{t^2}{3}=0,$ so $t=0$ or $3.$ These yield $x=0,3,-3,$ for three points. For $t+y=5,$ substitution gives $t^2-3t+6=0,$ which has no real root.

If $y<0,$ then $t>3.$ The equation $t-y=3$ gives $t^2+3t-18=0,$ whose only nonnegative root is the excluded boundary value $t=3.$ The equation $t-y=5$ gives $t^2+3t-24=0,$ with exactly one positive root $t=\frac{-3+\sqrt{105}}2>3.$ It yields two points, one for each sign of $x.$

The total number of ordered pairs is $3+2=5.$

Thus, D is the correct answer.