y2=81-x2=(9-x)(9+x) where 0≤x≤9 because a perfect square (y2 in this case) must be positive.
If y is an integer we just substitute all possible integer values for x and see if we get the perfect square of an integer:
x=9, y=0; x=8, y=√10; x=7, y=4√2; x=6, y=√45; x=5, y=2√14; x=4, y=√65; x=3, y=6√2; x=2, y=√77; x=1, y=4√5. The only integer value for y is zero (when x=9, or vice versa). But the question doesn't say that x and y have to be integers, so there are many possible solutions y=√(81-x2) for 0≤x≤9.