Answer
$$ y'=(2^x\ln 2+3^{-x}\ln 3)x^{-1} - (2^x-3^{-x} )x^{-2}.$$
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(a^x)'=a^x\ln{a}$
Since $ y=\frac{2^x-3^{-x}}{x}$, it is easier to simplify $ y $ as follows
$$ y=(2^x-3^{-x})x^{-1} $$
then, by the product rule, we have
$$ y'=(2^x\ln 2+3^{-x}\ln 3)x^{-1} - (2^x-3^{-x} )x^{-2}.$$
