\[\frac{1}{6x}\cdot\frac{6x+y}{y}=\frac{6x+y}{6xy}.\]
При \(x=\sqrt{32}=4\sqrt2\), \(y=\frac18\):
\[\frac{24\sqrt2+1/8}{3\sqrt2/4}=\frac{192\sqrt2+1}{6\sqrt2}=32+\frac{\sqrt2}{12}.\]
Ответ: \(32+\frac{\sqrt2}{12}\).