#### Answer

$$\frac{-e^5}{3}$$

#### Work Step by Step

Given
$$
T(x, y, z)=x e^{y-z}
$$
Consider $ P=(3,9,4) \text { and } Q=(5,7,3)$
Since $$\mathbf{v}=\overrightarrow{P Q}=\langle 5-3,7-9,3-4\rangle=\langle 2,-2,-1\rangle $$
Then the direction of $\mathbf{v}$ is
\begin{align*}
\mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\
&=\frac{\langle 2,-2,-1\rangle}{\sqrt{4+4+1}}\\
&=\left\langle\frac{2}{3},\frac{- 2}{3},\frac{- 1}{3}\right\rangle
\end{align*}
and
\begin{align*}
\nabla T&=\left\langle\frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z}\right\rangle\\
&=\left\langle e^{y-z}, x e^{y-z},-x e^{y-z}\right\rangle\\
\nabla T_{(3,9,4)}&=\left\langle e^5 ,3e^5, -3e^5\right\rangle \\
\end{align*}
Then the directional derivatives are given by
\begin{align*}
D_{\mathbf{u}} f(3,9,4)&=\nabla f_{(3,9,4)} \cdot \mathbf{u}\\
&=\left\langle e^5 ,3e^5, -3e^5\right\rangle \cdot\left\langle\frac{2}{3},\frac{- 2}{3},\frac{- 1}{3}\right\rangle\\
&=\frac{-e^5}{3}
\end{align*}