We know that equation of normal of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is
$$
\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2-b^2
$$

$\therefore$ Equations of normal to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ is
$$
\begin{aligned}
\frac{9 x}{x_1}+\frac{4 y}{y_1} & =9+4 \\
\Rightarrow \quad \frac{9 x}{x_1}+\frac{4 y}{y_1} & =13
\end{aligned}
$$
Since, line $x+y=k$ is normal to the given hyperbola
$$
\begin{aligned}
\therefore \quad \frac{9}{x_1} & =\frac{4}{y_1} \\
\Rightarrow \quad \frac{9}{x_1} & =\frac{4}{y_1} \\
& =\frac{13}{k} \\
\Rightarrow \quad x_1 & =\frac{9 k}{13} \\
y_1 & =\frac{4 k}{13}
\end{aligned}
$$


Since $\left(x_1, y_1\right)$ lie on the hyperbola.
$$
\begin{array}{rlrl}
& \therefore \frac{\left(\frac{9 k}{13}\right)^2}{9}-\frac{\left(\frac{4 k}{13}\right)^2}{4} & =1 \\
\Rightarrow \quad & \frac{9 k^2}{169}-\frac{4 k^2}{169} & =1 \\
\Rightarrow & & k k^2 & =169 \\
\Rightarrow & k & = \pm \frac{13}{\sqrt{5}}
\end{array}
$$