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There exists $f(x,y)=\dfrac{1}{\sqrt{16-x^2-y^2}}$ if and only if $16-x^2-y^2>0$ or equivalently if and only if $x^2+y^2<4^2$ (open disk $D((0,0),4)$). That is, $$\boxed{\:\mbox{Dom } (f)=D((0,0),4)\:}$$ On the other hand, for every $(x,y)$ such that $x^2+y^2=r^2$ with $0\leq r<4$ we have $f(x,y)=\dfrac{1}{\sqrt{16-r^2}}$ so, $f(x,y)$ varies from $1/4$ (included) to $+\infty$ (excluded). That is, $$\boxed{\:\mbox{Range } (f)=\left[1/4,+\infty\right)\:}$$