Let x=3+∆x when y=2+∆y, so 2+∆y=(4+∆x)^½=2(1+∆x/4)^½=2(1+∆x/8) when ∆x is very small.
Therefore 2+∆y=2+∆x/4, ∆y=∆x/4, ∆x=4∆y. When ∆y=0.0002, ∆x=0.0008.
CHECK
Let x=3.0008, then y=√4.0008=2.0002 approximately (more accurately y=2.00019999). So x is within 0.0008 of 3.