1. **Problem Statement:** Given a function $u = f(x^2 + 2yz, y^2 + 2xz)$, find the value of $$z \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial z} - x \frac{\partial u}{\partial y}.$$ 2. **Step 1: Define intermediate variables** Let $$s = x^2 + 2yz, \quad t = y^2 + 2xz.$$ Then, $u = f(s,t)$. 3. **Step 2: Use chain rule for partial derivatives** We have $$\frac{\partial u}{\partial x} = \frac{\partial f}{\partial s} \frac{\partial s}{\partial x} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial x},$$ $$\frac{\partial u}{\partial y} = \frac{\partial f}{\partial s} \frac{\partial s}{\partial y} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial y},$$ $$\frac{\partial u}{\partial z} = \frac{\partial f}{\partial s} \frac{\partial s}{\partial z} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial z}.$$ 4. **Step 3: Compute partial derivatives of $s$ and $t$** $$\frac{\partial s}{\partial x} = 2x, \quad \frac{\partial s}{\partial y} = 2z, \quad \frac{\partial s}{\partial z} = 2y,$$ $$\frac{\partial t}{\partial x} = 2z, \quad \frac{\partial t}{\partial y} = 2y, \quad \frac{\partial t}{\partial z} = 2x.$$ 5. **Step 4: Substitute into the expression** Calculate $$z \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial z} - x \frac{\partial u}{\partial y} = z \left( f_s 2x + f_t 2z \right) + y \left( f_s 2y + f_t 2x \right) - x \left( f_s 2z + f_t 2y \right),$$ where $f_s = \frac{\partial f}{\partial s}$ and $f_t = \frac{\partial f}{\partial t}$. 6. **Step 5: Expand and group terms** $$= 2x z f_s + 2z^2 f_t + 2y^2 f_s + 2x y f_t - 2x z f_s - 2x y f_t.$$ Notice that $2x z f_s$ and $-2x z f_s$ cancel out, and $2x y f_t$ and $-2x y f_t$ cancel out. 7. **Step 6: Simplify the remaining terms** $$= 2z^2 f_t + 2y^2 f_s.$$ **Final answer:** $$\boxed{2 y^2 \frac{\partial f}{\partial s} + 2 z^2 \frac{\partial f}{\partial t}}.$$