Multi variable function and its corresponding range? How do we find the range of
$z=f(x,y)=1/\ln(4−x^2−y^2)$
1) Replace $t=\ln(4−x^2−y^2)$ and $t\in (−\infty,\ln4)$ and there is a DNE at ln('')=0
2) Find range of $1/t$ ; How is this done?
Answers: $(−\infty,0)∪(1/\ln4,\infty)$

We know the domain of $f(x,y)=\dfrac{1}{\ln(4-x^2-y^2)}$ is
$$\color{blue}{D_f=\{(x,y)\in\mathbb{R}^2:x^2+y^2<4~,~x^2+y^2\neq3\}}$$
and in this domain $0\leq x^2+y^2<4$ so
$$0< 4-x^2-y^2\leq4~,~(x^2+y^2\neq3)$$
the function $\ln x$ is increasing then
$$-\infty< \ln(4-x^2-y^2)\leq\ln4~,~(x^2+y^2\neq3)$$
with reciprocating we have some irregular points for $\dfrac{1}{\ln(4-x^2-y^2)}$ in $x^2+y^2=3$, and except points on this circle we see
\begin{cases}
f(x,y)\in(-\infty,0)~~~\text{for}~~~0< 4-x^2-y^2<3, \\
f(x,y)\in(\dfrac{1}{\ln4},+\infty)~~~\text{for}~~~3< 4-x^2-y^2<4.
\end{cases}
so
$$\color{blue}{R_f=(-\infty,0)\cup(\dfrac{1}{\ln4},+\infty)}$$
