Step-by-step Solution:

To solve the system of equations for positive real numbers \(x\) and \(y\): \[ \begin{cases} x^2 + y \sqrt{xy} = 336 \\ y^2 + x \sqrt{xy} = 112 \end{cases} \] 1. Divide the First Equation by \( \sqrt{x} \): \[ \frac{x^2}{\sqrt{x}} + \frac{y \sqrt{xy}}{\sqrt{x}} = \frac{336}{\sqrt{x}} \] Simplify: \[ x \sqrt{x} + y \sqrt{y} = \frac{336}{\sqrt{x}} \quad \text{(1)} \] 2. Divide the Second Equation by \( \sqrt{y} \): \[ \frac{y^2}{\sqrt{y}} + \frac{x \sqrt{xy}}{\sqrt{y}} = \frac{112}{\sqrt{y}} \] Simplify: \[ y \sqrt{y} + x \sqrt{x} = \frac{112}{\sqrt{y}} \quad \text{(2)} \] 3. Set Equations (1) and (2) Equal: Since both expressions equal \( x \sqrt{x} + y \sqrt{y} \), we have: \[ \frac{336}{\sqrt{x}} = \frac{112}{\sqrt{y}} \] Simplify: \[ 3 \sqrt{y} = \sqrt{x} \quad \text{(3)} \] 4. Square Both Sides of Equation (3): \[ 9y = x \quad \text{(4)} \] 5. Substitute \( x = 9y \) into the First Original Equation: \[ (9y)^2 + y \sqrt{(9y) \cdot y} = 336 \] Simplify: \[ 81y^2 + y \cdot 3y = 336 \] \[ 81y^2 + 3y^2 = 336 \] \[ 84y^2 = 336 \] \[ y^2 = 4 \Rightarrow y = 2 \] 6. Find \( x \) Using Equation (4): \[ x = 9y = 9 \cdot 2 = 18 \] 7.Calculate \( x + y \): \[ x + y = 18 + 2 = 20 \] Therefore, the value of \( x + y \) is: \[ \boxed{20} \]