Let y=½ln((x+3)/(x-3))=½[ln(x+3)-ln(x-3)] (I have assumed that log is the same as ln.)
y⁽¹⁾=½[1/(x+3)-1/(x-3)]
y⁽²⁾=½[-1/(x+3)²+1/(x-3)²]
y⁽³⁾=½[2/(x+3)³-2/(x-3)³]
y⁽⁴⁾=½[-3!/(x+3)⁴+3!/(x-3)⁴]
...
y⁽ⁿ⁾=½(-1)ⁿ(n-1)![-1/(x+3)ⁿ+1/(x-3)ⁿ] or ½(-1)ⁿ(n-1)![1/(x-3)ⁿ-1/(x+3)ⁿ].