= Solution
The <inverse function theorem> states that a $C^1$ map on an open subset of $\mathbb R^2$ has a $C^1$ local inverse near a point where its derivative is invertible; the inverse derivative at the image point is the inverse <matrix> of the original derivative.
Put $s=1+x+y$. Quotient differentiation gives
$$
\boxed{Df(x,y)=\frac1{s^2}\begin{pmatrix}1+y&-x\\-y&1+x\end{pmatrix},
\qquad\det Df(x,y)=\frac1{s^3}.}
$$
The entries are smooth and the determinant is nonzero everywhere on the stated domain, so the theorem supplies the required local inverse at every point. The point $(1/3,1/3)$ is the image of $(1,1)$, where
$$
Df(1,1)=\frac19\begin{pmatrix}2&-1\\-1&2\end{pmatrix}.
$$
Consequently
$$
\boxed{Df^{-1}(1/3,1/3)=\begin{pmatrix}6&3\\3&6\end{pmatrix}.}
$$
Directly solving the two rational equations also gives $f^{-1}(u,v)=(u/(1-u-v),v/(1-u-v))$ on $u+v\neq1$, consistent with the derivative calculation.
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