To find the directional derivative $D_{\mathbf{u}} f(x, y)$ of $f(x, y) = e^x \sin y$ at the point $P\left(0, \frac{\pi}{3}\right)$ in the direction of the vector $\mathbf{v} = -6\mathbf{i} + 8\mathbf{j}$, we will follow these steps:

Step 1: Compute the Gradient of $f(x, y)$

The gradient of $f(x, y)$, denoted as $\nabla f(x, y)$, is: $\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$ For $f(x, y) = e^x \sin y$:

1. Partial derivative with respect to $x$: $\frac{\partial f}{\partial x} = e^x \sin y$
2. Partial derivative with respect to $y$: $\frac{\partial f}{\partial y} = e^x \cos y$

Thus, $\nabla f(x, y) = \left( e^x \sin y, e^x \cos y \right)$

Step 2: Evaluate $\nabla f(x, y)$ at $P\left(0, \frac{\pi}{3}\right)$

Substitute $x = 0$ and $y = \frac{\pi}{3}$ into the gradient:

1. For $\frac{\partial f}{\partial x}$: $e^0 \sin\left(\frac{\pi}{3}\right) = 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$
2. For $\frac{\partial f}{\partial y}$: $e^0 \cos\left(\frac{\pi}{3}\right) = 1 \cdot \frac{1}{2} = \frac{1}{2}$

So, $\nabla f\left(0, \frac{\pi}{3}\right) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)$.

Step 3: Normalize the Direction Vector $\mathbf{v}$

The direction vector $\mathbf{v} = -6\mathbf{i} + 8\mathbf{j}$ needs to be converted into a unit vector $\mathbf{u}$.

1. Magnitude of $\mathbf{v}$: $\|\mathbf{v}\| = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$
2. Unit vector $\mathbf{u}$: $\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \frac{-6\mathbf{i} + 8\mathbf{j}}{10} = -0.6\mathbf{i} + 0.8\mathbf{j}$

Step 4: Compute the Directional Derivative

The directional derivative of $f$ at $P\left(0, \frac{\pi}{3}\right)$ in the direction of $\mathbf{u}$ is: $D_{\mathbf{u}} f(x, y) = \nabla f(x, y) \cdot \mathbf{u}$ At $P\left(0, \frac{\pi}{3}\right)$: [ D_{\mathbf{u}} f\left(0, \frac{\pi}{3}\right) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \cdot \left( -0.6, 0