To solve the problem, we need to calculate $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z}$ given that:

$u = \frac{y}{x} + \frac{z}{x}.$

Step 1: Rewrite $u$ in a simpler form

We can express $u$ as: $u = \frac{y + z}{x}.$

Step 2: Find the partial derivatives of $u$ with respect to $x$, $y$, and $z$.

1. Partial derivative with respect to $x$: $\frac{\partial u}{\partial x} = \frac{\partial}{\partial x} \left( \frac{y + z}{x} \right) = -\frac{y + z}{x^2}.$

2. Partial derivative with respect to $y$: $\frac{\partial u}{\partial y} = \frac{\partial}{\partial y} \left( \frac{y + z}{x} \right) = \frac{1}{x}.$

3. Partial derivative with respect to $z$: $\frac{\partial u}{\partial z} = \frac{\partial}{\partial z} \left( \frac{y + z}{x} \right) = \frac{1}{x}.$

Step 3: Substitute into $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z}$

Now, we calculate each term individually:

1. $x \frac{\partial u}{\partial x} = x \cdot \left( -\frac{y + z}{x^2} \right) = -\frac{y + z}{x}.$

2. $y \frac{\partial u}{\partial y} = y \cdot \frac{1}{x} = \frac{y}{x}.$

3. $z \frac{\partial u}{\partial z} = z \cdot \frac{1}{x} = \frac{z}{x}.$

Adding these together:

$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = -\frac{y + z}{x} + \frac{y}{x} + \frac{z}{x}.$

Simplify the expression:

$= -\frac{y + z}{x} + \frac{y + z}{x} = 0.$

Final Answer:

$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = 0.$

Would you like more details, or do you have any questions?

Here are some related questions:

1. How do you find partial derivatives for more complex functions?
2. What is the significance of calculating $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z}$?
3. Can we apply this method to functions involving exponential or trigonometric terms?
4. How does the result change if $u$ is defined as a different ratio?
5. What are common applications of such derivative calculations in physics or engineering?

Tip: Simplifying expressions early can make calculations with partial derivatives more manageable.