Past exam of the options I course of the University of Oxford / 2010 / F / F1

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 givesThe 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 $(1,1)$, whereConsequentlyDirectly solving the two rational equations also gives $f^{-1}(u,v)=(u/(1-u-v),v/(1-u-v))$ on $u+v\ne 1$, consistent with the derivative calculation.

For each fixed $(x,y)$, the function $z\mapsto z^3+z+xy-1$ is continuous, strictly increasing because its derivative is $3z^2+1>0$, and tends to opposite infinities at the two ends of the real line. The intermediate value theorem therefore gives exactly one zero.

For $F(x,y,z)=z^3+z+xy-1$, its partial derivative $F_z=3z^2+1$ never vanishes. The implicit function theorem gives a local $C^1$ solution near every point of the solution set. Uniqueness makes these local solutions agree, producing a global $g\in C^1({\mathbb{R}}^2)$. Implicit differentiation givesAt $(1,1)$ the root is $g(1,1)=0$, so