1. Given that the real number $x$ satisfies $\sqrt{x^{2}-x^{3}}=x$. $\sqrt{1-x}$. Then the range of values for $x$ is
2.15.

From $2 x^{2}-6 x+y^{2}=0$ we get
$$
y^{2}=-2 x^{2}+6 x \geqslant 0 \text {. }
$$

Solving this, we get $0 \leqslant x \leqslant 3$.
Let $z=x^{2}+y^{2}+2 x$. Then
$$
\begin{array}{l} 
z=x^{2}-2 x^{2}+6 x+2 x \\
=-x^{2}+8 x \\
=-(x-4)^{2}+16 . \\
\because 0 \leqslant x \leqslant 3,
\end{array}
$$
$\therefore$ when $x=3$, $z$ has a maximum value of 15.