Slope of normal $=-\left(\frac{d x}{d y}\right)_{\left(x_1, y_1\right)}$
Given curve, $y=\frac{x}{1+x^2}$, then
$$
\begin{aligned}
\frac{d y}{d x} & =\frac{\left(1+x^2\right)-x(2 x)}{\left(1+x^2\right)^2}=\frac{1-x^2}{\left(1+x^2\right)^2} \\
\Rightarrow \quad-\frac{d x}{d y} & =-\frac{\left(1+x^2\right)^2}{1-x^2}=\frac{\left(1+x^2\right)^2}{x^2-1}
\end{aligned}
$$
At $x=-4$
$$
\begin{aligned}
\left(-\frac{d x}{d y}\right)_{x=-4} & =\frac{\left((-4)^2+1\right)^2}{(-4)^2-1} \\
& =\frac{(16+1)^2}{16-1}=\frac{17^2}{15}=\frac{289}{15}
\end{aligned}
$$
$\therefore$ Slope of normal to curve $y=\frac{x}{x^2+1}$, at $x=-4$
is $289 / 15$.