Answer
$$\frac{2 \ln(2e)}{\sqrt{6}}$$
Work Step by Step
Given
$$g(x, y, z)=x \ln (y+z), \quad \mathbf{v}=2 \mathbf{i}-\mathbf{j}+\mathbf{k}, \quad P=(2, e, e)
$$
Since the direction of $\mathbf{v}$ is
\begin{align*}
\mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\
&=\frac{\langle 2,-1,1\rangle}{\sqrt{4+1+1}}\\
&=\left\langle\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}}\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 \ln (y+z), \frac{x}{y+z},\frac{x}{y+z}\right\rangle \\
\nabla g_{(2,e,e)}&=\left\langle \ln(2e), \frac{1}{e},\frac{1}{e}\right\rangle
\end{align*}
Then the directional derivatives are given by
\begin{align*}
D_{\mathbf{u}} g(2,e,e)&=\nabla g_{(2,e,e)} \cdot \mathbf{u}\\
&=\left\langle \ln(2e), \frac{1}{e},\frac{1}{e}\right\rangle \cdot \left\langle\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}}\right\rangle\\
&=\frac{2 \ln(2e)}{\sqrt{6}}
\end{align*}
