Answer
$$\frac{-1}{\sqrt{3}}$$
Work Step by Step
Given
$$g(x, y, z)=x e^{-y z}, \quad \mathbf{v}=\langle 1,1,1\rangle, \quad P=(1,2,0)
$$
Since the direction of $\mathbf{v}$ is
\begin{align*}
\mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\
&=\frac{\langle 1,1,1\rangle}{\sqrt{1+1+1}}\\
&=\left\langle\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right\rangle
\end{align*}
and
\begin{align*}
\nabla g&=\left\langle\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z}
\right\rangle\\
&=\left\langle e^{-y z},-x z e^{-y z},-x y e^{-y z}\right\rangle \\
\nabla g_{(1,2,0)}&=\left\langle 1, 0,-2\right\rangle
\end{align*}
Then the directional derivatives are given by
\begin{align*}
D_{\mathbf{u}} g(1,2,0)&=\nabla g_{(1,2,0)} \cdot \mathbf{u}\\
&=\left\langle 1, 0,-2\right\rangle \cdot\left\langle\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right\rangle\\
&=\frac{-1}{\sqrt{3}}
\end{align*}
