$g(x)=\sqrt{f(x)}\Rightarrow {{D}_{g}}=\left[ -2,1 \right]$  $h(x)=\frac{1}{\sqrt{2x-{{x}^{2}}}}\Rightarrow {{D}_{h}}=\left\{ x\left| 2x-{{x}^{2}} \gt 0 \right. \right\}=(0,2)$  ${{D}_{m}}={{D}_{g}}\cap {{D}_{h}}=\left[ -2,1 \right]\cap (0,2)=\left( 0,1 \right]$