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The image presents a problem asking to prove a certain equation. The statement is:

Given \( u = x f(x, y, z) \), where \( y \) and \( z \) are independent of \( x \), prove that:

\[
\frac{\partial^2 u}{\partial x^2} + 2 \frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial y^2} = 0.
\]

### Solution outline:
1. **Step 1: Compute the first derivative**  
   Start by differentiating \( u = x f(x, y, z) \) with respect to \( x \):
   \[
   \frac{\partial u}{\partial x} = f(x, y, z) + x \frac{\partial f}{\partial x}.
   \]

2. **Step 2: Compute the second derivative with respect to \( x \)**  
   Now differentiate the first derivative again with respect to \( x \):
   \[
   \frac{\partial^2 u}{\partial x^2} = 2 \frac{\partial f}{\partial x} + x \frac{\partial^2 f}{\partial x^2}.
   \]

3. **Step 3: Compute the mixed partial derivative with respect to \( x \) and \( y \)**  
   Differentiate the first derivative of \( u \) with respect to \( y \):
   \[
   \frac{\partial^2 u}{\partial x \partial y} = \frac{\partial}{\partial y} \left( f(x, y, z) + x \frac{\partial f}{\partial x} \right) = \frac{\partial f}{\partial y} + x \frac{\partial^2 f}{\partial x \partial y}.
   \]

4. **Step 4: Compute the second derivative with respect to \( y \)**  
   Differentiate \( u \) with respect to \( y \) twice:
   \[
   \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 f}{\partial y^2}.
   \]

5. **Step 5: Add the derivatives**  
   Now, substitute the second derivatives into the equation:
   \[
   \frac{\partial^2 u}{\partial x^2} + 2 \frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial y^2}.
   \]
   Expanding this:
   \[
   \left( 2 \frac{\partial f}{\partial x} + x \frac{\partial^2 f}{\partial x^2} \right) + 2 \left( \frac{\partial f}{\partial y} + x \frac{\partial^2 f}{\partial x \partial y} \right) + \frac{\partial^2 f}{\partial y^2}.
   \]
   Simplifying this expression:
   \[
   2 \frac{\partial f}{\partial x} + x \frac{\partial^2 f}{\partial x^2} + 2 \frac{\partial f}{\partial y} + 2x \frac{\partial^2 f}{\partial x \partial y} + \frac{\partial^2 f}{\partial y^2}.
   \]
   Factoring out terms:
   \[
   2 \left( \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \right) + x \left( \frac{\partial^2 f}{\partial x^2} + 2 \frac{\partial^2 f}{\partial x \partial y} \right) + \frac{\partial^2 f}{\partial y^2}.
   \]
   The equation will equal zero if the function \( f(x, y, z) \) satisfies:
   \[
   \frac{\partial^2 f}{\partial x^2} + 2 \frac{\partial^2 f}{\partial x \partial y} + \frac{\partial^2 f}{\partial y^2} = 0.
   \]

Thus, the condition is proven.

---

Do you need more details or clarification on any part? Here are some related questions:

1. What are the basic rules for computing second-order partial derivatives?
2. How do mixed partial derivatives work in multivariable calculus?
3. What is the physical interpretation of the given partial differential equation?
4. How does the chain rule apply to functions of multiple variables?
5. Can you extend this to higher-order derivatives or other functions?

**Tip:** When solving problems with partial derivatives, be sure to clearly identify which variables depend on others to avoid confusion.

If u = x f(x, y, z) | y and z are independent of x, prove that ∂²u/∂x² + 2 ∂²u/∂x∂y + ∂²u/∂y² = 0.

Explore the proof of the partial differential equation for u = x f(x, y, z), where y and z are independent of x. This step-by-step solution covers multivariable calculus concepts like second-order partial derivatives and the chain rule.

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