Question: The ellipse $E:=\lbrace(x,y) \in \mathbb{R^2}:x^2+y^2+xy-x-y \leq 1\rbrace$ is given. 

How to compute the area/Lebesgue measure of this set? Also, how an affine transformation $F: \mathbb{R^2} \to \mathbb{R^2}$ can be found such that $F(\lbrace (x,y) \in \mathbb{R^2}:x^2+y^2 \leq1 \rbrace)=E$?

I know that the area of an ellipse is $\pi \cdot $A$\cdot$B, but in this case the values of A, B aren't given.

To find an affine transformation I tried to use that such transformations on $\mathbb{R^2}$ are given by $T((x,y)^T)= A \cdot (x,y)^T+z$ for $A \in \mathbb{R^{2x2}}, z \in \mathbb{R^2}$. 

If $(0,0)$ maps to the center, $(1,0)$ and $(0,1)$ to the right vertex and top co-vortex, I get with the formula:

$T((0,0)^T)=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot (0,0)^T+z=z$

$T((1,0)^T)=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot (1,0)^T+z=(a,c)^T+z$

$T((0,1)^T)=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot (0,1)^T+z=(b,a)^T+z$

This looks wrong to me, since I should get a linear system with six equations which can be solved.

So what exactly has to be done to determine the area of this ellipse and find the affine transformation, that maps the unit circle to this ellipse?

Answer: 
Let $A\subset[0,1]$ be a closed set of positive measure that doesn't contain any interval ([this](https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set), for instance). Then the function $\phi(x)=d(x,A)$ is Lipshchitz, hence absolutely continuous, and $\{\phi=0\}=A$ doesn't contain any interval.
No, any closed subset of $[0,1]$ can be the zero set of a $C^\infty$ function, and such subsets can have positive measure but no interior (e.g. a fat Cantor set).