Domain, range and zeros of $f(x,y)=\frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}$ Given the following function with two variables:
\begin{equation}
    \frac{\sqrt{4x-x^2-y^2}}{x^2y^2-4xy^2+3y^2}
\end{equation}
I need to find a) the domain for the above function.
Can anyone give me a hint on how to find the domain in f?
I already know that:
\begin{equation}
    dom f = \{ f(x,y) ∈ \mathbb{R}^2\backslash(x^2y^2-4xy^2+3y^2=0) : \sqrt{4x-x^2-y^2} \geq 0 \}
\end{equation}
But of course this needs to be written in a more simpler form.
During class we solve simpler functions  like without fractions and roots, so I don't have anything that can help me.
After this I also need to find:


*

*b) zeros of the function  

*c) Calculating Algebraically, the range of the function:
\begin{equation}
T(x,y) = \sqrt{4x-x^2-y^2}
\end{equation}

*d) Extreme values of the function T


I'm of course not expecting the complete solution but something like a kick start.

we have for the domian two cases
$4x-x^2-y^2\geq 0$ and $y^2(x^2-4x+3)>0$
or
$4x-x^2-y^2\le 0$ and $y^2(x^2-4x+3)<0$
note that $x^2-4x+3=0$ gives $x_1=1$ or $x_2=3$ and $4x-x^2-y^2=4-(y^2+(x-2)^2)$ thus we get for the first case
$4>y^2+(x-2)^2$ and $x>3$ and $y\neq 0$
or
$4>y^2+(x-2)^2$ and $x<1$ and $y \neq 0$
and for the second case
$(x-2)^2+y^2\geq 4$ and $y\neq 0$ and $1<x<3$
sorry i thougt the problem was  $\sqrt{\frac{4x-x^2-y^2}{x^2y^2-4xy^2+3y^2}}$
b) the zeros of our functions are $(x-2)^2+y^2=4$ and $y^2\neq 0$ and $x\neq 3$ and $x\neq 1$
c) we have to solve
$T_x=\frac{2-x}{\sqrt{4x-x^2-y^2}}=0$
and
$T_y=\frac{y}{\sqrt{4x-x^2-y^2}}=0$
a) for the domain we have $(x-2)^2+y^2\le 4$ and $y\neq 0$ and $x\neq 3$ and $x\neq 1$
