Solution
= Solution
Let $P=\mathbf z\mathbf z^T$, so $P^2=r^2P$ and $f^2=1+r^2$. Multiplying the proposed <inverse matrix> by the <Hessian> gives
$$
\left(\frac I f-\frac P{f^3}\right)f(I+P)=I+P-\frac{P+P^2}{f^2}=I.
$$
Hence \b[$\displaystyle H_f^{-1}=f(I+\mathbf z\mathbf z^T)=f\begin{pmatrix}1+x^2&xy\\xy&1+y^2\end{pmatrix}$], as required. Its existence also follows from the <Hessian> being a <positive-definite matrix>.