Answer
$$y' = \frac{4}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} \cr
& {\text{use quotient rule }} \cr
& \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr
& y' = \frac{{\left( {{e^x} + {e^{ - x}}} \right)\left[ {{e^x} - {e^{ - x}}} \right]' - \left( {{e^x} - {e^{ - x}}} \right)\left[ {{e^x} + {e^{ - x}}} \right]'}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cr
& y' = \frac{{\left( {{e^x} + {e^{ - x}}} \right)\left( {{e^x} + {e^{ - x}}} \right) - \left( {{e^x} - {e^{ - x}}} \right)\left( {{e^x} - {e^{ - x}}} \right)}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cr
& {\text{multiply}} \cr
& y' = \frac{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2} - {{\left( {{e^x} - {e^{ - x}}} \right)}^2}}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cr
& {\text{simplify}} \cr
& y' = \frac{{{e^{2x}} + 2{e^{x - x}} + {e^{ - 2x}} - {e^{2x}} + 2{e^{x - x}} - {e^{ - 2x}}}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cr
& y' = \frac{{{e^{2x}} + 2 + {e^{ - 2x}} - {e^{2x}} + 2 - {e^{ - 2x}}}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cr
& y' = \frac{4}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cr} $$