Since $\sqrt{x}$ = integer, the value of $x$ should be a perfect square. At the same time, the value of $y$ should be such that its square when multiplied by $x$ gives $36$

The possible combinations are

$x \times y^2 = 36$

$1 \times 36 => x=1, \sqrt{x}=1, y^2=36, y=6 \text{ or } -6$

$4 \times 9 => x=4, \sqrt{x}=2, y^2=9, y=3 \text{ or } -3$

$9 \times 4 => x=9, \sqrt{x}=3, y^2=4, y=2 \text{ or } -2$

$36 \times 1 => x=36, \sqrt{x}=6, y^2=1, y=1 \text{ or } -1$

Therefore the possible values of $y$ are $1,-1,2,-2,3,-3,6,-6$.

There are Eight possible values of $y$.

The answer is E.