= Solution
{c}
The <inverse function theorem> says that a continuously differentiable map on an open set, with invertible derivative at a point, restricts to a $C^1$ diffeomorphism between neighborhoods of that point and its image. The inverse derivative is the inverse <matrix> of the original derivative at the corresponding preimage.
Put $s=1+x+y\ne0$. Differentiating each coordinate gives
$$
\boxed{Df(x,y)=s^{-2}\begin{pmatrix}1+y&-x\\-y&1+x\end{pmatrix},\qquad\det Df=s^{-3}.}
$$
The determinant never vanishes on $\Omega$, so the theorem applies at every point. The image point $(1/3,1/3)$ has preimage $(1,1)$: the equations $x/s=y/s=1/3$ imply $x=y=1$. Therefore
$$
\boxed{Df^{-1}(1/3,1/3)=\left[\frac19\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\right]^{-1}
=\begin{pmatrix}6&3\\3&6\end{pmatrix}.}
$$
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