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.

Hints-kick start:


*

*$4x-x^2-y^2=4-(x-2)^2-y^2$

*$(x-2)^2+y^2=4$ is the equation of a circle

*$x^2y^2-4xy^2+3y^2=y^2((x-2)^2-1)=y^2(x-3)(x-1)$

