Solution
= Solution
For <Newton's method in optimization>, let $\Delta=H(f)^{-1}\nabla f$. Its second row equation is $-4\Delta_x+4\Delta_y=4y-4x$, so $\Delta_y=\Delta_x+y-x$. Therefore $y_{n+1}=y_n-\Delta_y=x_n-\Delta_x=x_{n+1}$. The given first coordinate yields
$$
\boxed{\varphi(x,y)=\frac{2x^3}{3x^2-1}.}
$$
The <Newton step> is defined where $\det H(f)=16(3x^2-1)\ne0$; both coordinates after any defined step are equal.