To find the joint density $f_{XY}(x, y)$ of $(X, Y)$ given the specified region $D$, we follow these steps:

1. Identify the region $D$: The region $D$ is given by: $D = \{(x, y) \mid 3x > y, 0 < x < 1, y > 0\}$

   This implies that $y$ must be less than $3x$ for $0 < x < 1$.

2. Determine the area of region $D$: Since $X$ and $Y$ are uniformly distributed over the region $D$, we need to compute the area of $D$ to find the joint density. The region is bounded by the lines $y = 0$, $y = 3x$, $x = 0$, and $x = 1$.

   The area of this triangular region can be calculated as: $\text{Area}(D) = \int_{0}^{1} \int_{0}^{3x} dy \, dx$ Evaluate the inner integral first: $\int_{0}^{3x} dy = 3x$ Now, integrate with respect to $x$: $\int_{0}^{1} 3x \, dx = 3 \int_{0}^{1} x \, dx = 3 \left[ \frac{x^2}{2} \right]_{0}^{1} = 3 \cdot \frac{1}{2} = \frac{3}{2}$

   Therefore, the area of $D$ is $\frac{3}{2}$.

3. Find the joint density: Since $(X, Y)$ is uniformly distributed over $D$, the joint density function $f_{XY}(x, y)$ is constant within $D$ and zero outside of $D$. The constant value is the reciprocal of the area of $D$.

   Thus, the joint density $f_{XY}(x, y)$ is:

   \frac{2}{3} & \text{if } 0 < x < 1 \text{ and } 0 < y < 3x \\ 0 & \text{otherwise} \end{cases}$$

Therefore, the correct joint density function is: