${{D}_{\frac{2f-g}{f.g}}}={{D}_{f}}\bigcap {{D}_{g}}-\left\{ x\left| (f.g)(x)=0 \right. \right\}=\left\{ 3,4,5 \right\}-\left\{ 3 \right\}=\left\{ 4,5 \right\}$ $(\frac{2f-g}{f.g})(4)=\frac{2f(4)-g(4)}{f(4)g(4)}=\frac{2\times 2-(-1)}{2\times (-1)}=\frac{5}{-2}=-2/5$ $(\frac{2f-g}{f.g})(5)=\frac{2f(5)-g(5)}{f(5)g(5)}=\frac{2\times 3-(-3)}{3\times (-3)}=-1$ $\Rightarrow \frac{2f-g}{f.g}=\left\{ (4,-2/5),(5,-1) \right\}$