15. Given the parabola $y=x^{2}$ and the line $l$ passing through the point $P(-1,-1)$ intersects the parabola at points $P_{1}$ and $P_{2}$. Find:
(1) The range of the slope $k$ of the line $l$;
(2) The locus of point $Q$ on the segment $P_{1} P_{2}$ that satisfies the condition
$$
\frac{1}{P P_{1}}+\frac{1}{P P_{2}}=\frac{2}{P Q}
$$
15. (1) The line $l: y+1=k(x+1)$ intersects with the parabola equation $y=x^{2}$. By eliminating $y$ and rearranging, we get
$$
x^{2}-k x-(k-1)=0 \text {. }
$$

From $\Delta=(-k)^{2}+4(k-1)>0$, solving gives
$k>-2+2 \sqrt{2}$ or $k<-2-2 \sqrt{2}$.
\end{array}
$$

Thus, $x_{1}+1, x_{2}+1, x+1$ have the same sign.
$$
\begin{array}{l}
\text { Then } \frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}=\frac{2}{x+1} \\
\Rightarrow x=\frac{4}{x_{1}+x_{2}+2}-1=\frac{2-k}{k+2} \\
\Rightarrow k=\frac{2-2 x}{x+1} \\
\Rightarrow y=k(x+1)-1=1-2 x \\
\Rightarrow 2 x-y+1=0 .
\end{array}
$$

Since $k>-2+2 \sqrt{2}$ or $k<-2-2 \sqrt{2}$, it follows that $-\sqrt{2}-1<x<\sqrt{2}-1$, and $x \neq-1$.
Therefore, the trajectory equation of point $Q$ is
$$
2 x-y+1=0 \text {, }
$$

where $-\sqrt{2}-1<x<\sqrt{2}-1$, and $x \neq-1$.