If the square root of \((63 - 3\sqrt{x})\) is an integer, then it has to be a perfect square.
Here are perfect squares under 63:
49 = 7^2
36 = 6^2
25 = 5^2
16 = 4^2
9 = 3^2
4 = 2^2
1 = 1^2
0 = 0^2
Since there are 8 perfect squares under 63, then there are 8 real values of \(x\) where the \(\sqrt{(63 - 3\sqrt{x})}\) is an integer.