#### Answer

$$0$$

#### Work Step by Step

Given
$$
f(x, y)=\sin (x-y), \quad \mathbf{v}=\langle 1,1\rangle, \quad P=\left(\frac{\pi}{2}, \frac{\pi}{6}\right)
$$
Since the direction of $\mathbf{v}$ is
\begin{align*}
\mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\
&=\frac{\langle 1,1\rangle}{\sqrt{1+1}}\\
&=\left\langle\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\rangle
\end{align*}
and
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle\\
&=\left\langle \cos (x-y), -\cos (x-y)\right\rangle \\
\nabla f_{(\pi/2,\pi/6)}&=\left\langle \frac{1}{2}, \frac{-1}{2}\right\rangle
\end{align*}
Then the directional derivatives are given by
\begin{align*}
D_{\mathbf{u}} f(\pi/2,\pi/6)&=\nabla f_{(\pi/2,\pi/6)} \cdot \mathbf{u}\\
&=\left\langle \frac{1}{2}, \frac{-1}{2}\right\rangle \cdot\left\langle\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\rangle\\
&=0
\end{align*}