= Solution
For the <vector> equation, the <Jacobian matrix> is
$$
J(x,y)=
\begin{pmatrix}
6x^5&-20y^3\\1&-1
\end{pmatrix}.
$$
Write the updated coordinates as $(X,Y)$. The second equation of the <multivariable Newton iteration> gives $X-Y=0$. Substituting $X=Y$ into its first linearized equation gives
$$
(6x^5-20y^3)X=5x^6-15y^4-6.
$$
Hence \b[$h(x,y)=6x^5-20y^3$], and, whenever $h\ne0$,
$$
\boxed{
(x_{k+1},y_{k+1})
=\frac{5x_k^6-15y_k^4-6}{6x_k^5-20y_k^3}(1,1).
}
$$
Starting from $(0,1)$ gives $(x_1,y_1)=(21/20,21/20)$. Every subsequent iterate remains on the diagonal, where this is exactly the scalar <Newton root-finding iteration> from (b). The value $f(21/20)>0$, so $21/20<r$.
For $1\le x<r$, concavity again puts the tangent's zero at or above $r$. Also
$$
N'(x)=\frac{f(x)f''(x)}{f'(x)^2}<0,
$$
so $r\le N(21/20)\le N(1)=8/7<6/5$. After one scalar step the sequence therefore enters the invariant <interval> proved in (b). It converges to \b[$(r,r)$ at least quadratically]. If the <vector> error is measured in the <Euclidean norm>, then $\|(x_k,y_k)-(r,r)\|=\sqrt2\,|x_k-r|$, so the scalar quadratic estimate also gives a quadratic <vector> estimate. The initial off-diagonal step causes no difficulty because its <Jacobian matrix> is invertible and lands exactly on the diagonal.
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