Answer
\[
\left.(a) \frac{\partial z}{\partial x}\right|_{(x=3, z=\pm 2 \sqrt{6})}=\pm \frac{3}{2 \sqrt{6}}
\]
Work Step by Step
We are given that
\[
1=x^{2}+y^{2}-z^{2}
\]
Differentiating the changing variable $z$ with respect to $x$ and taking $y$ as constant
\[
\begin{aligned}
\frac{\partial}{\partial x}\left(-z^{2}+y^{2}+x^{2}\right) &=\frac{d}{d x}(1) \\
0=0+2 x-2 z \frac{\partial z}{\partial x} & \\
\frac{x}{z} =\frac{\partial z}{\partial x} & \\
\left.\frac{\partial z}{\partial x}\right|_{(x=3, z=\pm 2 \sqrt{6})} &=\pm \frac{3}{2 \sqrt{6}}
\end{aligned}
\]
First, we have to get the value of z and then differentiate it
\[
\begin{aligned}
\because 1=-z^{2}+y^{2}+x^{2} & \\
\Rightarrow z &=\pm \sqrt{-1+y^{2}+x^{2}}
\end{aligned}
\]
Differentiating the sides with respect to x with y constant
\[
\begin{aligned}
\frac{\partial z}{\partial x} &=\frac{\partial}{\partial x}(\pm \sqrt{x^{2}+y^{2}-1}) \frac{\partial z}{\partial x}=\frac{d}{d x}(\pm \sqrt{x^{2}+y^{2}-1}) \\
\frac{\partial z}{\partial x} &=\pm \frac{1}{2 \sqrt{x^{2}+y^{2}-1}} *(2 x) \\
\frac{\partial z}{\partial x} &=\pm \frac{x}{\sqrt{x^{2}+y^{2}-1}} \\
\left.\frac{\partial z}{\partial x}\right|_{(x=3, y=4)} &=\pm \frac{3}{\sqrt{3^{2}+4^{2}-1}} \\
\left.\frac{\partial z}{\partial x}\right|_{(x=3, y=4)} &=\pm \frac{3}{2 \sqrt{6}}
\end{aligned}
\]
