$\text{The sum of two numbers}x\text{and}y\text{(where}x\text{is greater than}y\text{) is equal to four times of their difference. Find the value of }\frac{5y}{3(x^2-y^2)}.$

The correct answer is $\frac{25}{16}$.

Step 1: Formulate the relationship between x and y.
We are given that the sum of two numbers $x$ and $y$ (where $x>y$) is equal to 4 times their difference:
$x+y=4(x-y)$

Expanding the right side of the equation:
$x+y=4x-4y$

Rearranging the terms to group $x$ and $y$ on opposite sides:
$y+4y=4x-x$
$5y=3x$
$\frac{x}{y}=\frac{5}{3}$

Step 2: Express x and y in terms of a constant k.
From the ratio $\frac{x}{y}=\frac{5}{3}$, we can write:
$x=5k$ and $y=3k$ (where $k$ is a non-zero constant).

Step 3: Evaluate the required expression.
We need to calculate the value of:
$\frac{5xy}{3(x^2-y^2)}$

First, calculate the numerator:
$\text{Numerator}=5\left(5k\right)\left(3k\right)=75k^2$

Next, calculate the denominator:

Now, divide the numerator by the denominator:
$\frac{75k^2}{48k^2}=\frac{75}{48}$

Simplifying the fraction by dividing both numerator and denominator by 3:
$\frac{75}{48}=\frac{25}{16}$

Thus, the value of the expression is $\frac{25}{16}$.