(9) (15 points) On the ellipse $x^{2}+4 y^{2}=4 x$, find the coordinates of the point $P$ that makes $z=x^{2}-y^{2}$ attain its maximum and minimum values.
(9. Let $P(x, y)$ be a point on the ellipse $x^{2}+4 y^{2}=4 x$, then $y^{2}=\frac{1}{4}(4 x-$
$$
\begin{aligned}
\left.x^{2}\right) \text {, substituting into } z & =x^{2}-y^{2}, \text { we get } \\
z & =x^{2}-\frac{1}{4}\left(4 x-x^{2}\right)=\frac{5}{4}\left(x-\frac{2}{5}\right)^{2}-\frac{1}{5} .
\end{aligned}
$$

Since $P(x, y)$ is a point on $x^{2}+4 y^{2}=4 x$, we have $4 x-x^{2}=4 y^{2} \geqslant 0$, so $0 \leqslant x \leqslant 4$.
Hence, when $\left\{\begin{array}{l}x=\frac{2}{5}, \\ y= \pm \frac{3}{5}\end{array}\right.$, $z$ has a minimum value of $-\frac{1}{5}$, at this time $P\left(\frac{2}{5}, \pm \frac{3}{5}\right)$;
when $\left\{\begin{array}{l}x=4, \\ y=0\end{array}\right.$, $z$ has a maximum value of 16, at this time $P(4,0)$.