We are asked to find the gradient of the function $f(x, y, z) = xz \ln(x + y + z)$. The gradient is given by $\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)$.

1. Problem Description

We are asked to find the gradient of the function $f(x, y, z) = xz \ln(x + y + z)$.

The gradient is given by $\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)$.

2. Solution Steps

First, we find the partial derivative of $f$ with respect to $x$:

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(xz \ln(x + y + z))$

Using the product rule, we have

$\frac{\partial f}{\partial x} = z \ln(x + y + z) + xz \cdot \frac{1}{x + y + z} \cdot 1 = z \ln(x + y + z) + \frac{xz}{x + y + z}$

Next, we find the partial derivative of $f$ with respect to $y$:

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(xz \ln(x + y + z))$

$\frac{\partial f}{\partial y} = xz \cdot \frac{1}{x + y + z} \cdot 1 = \frac{xz}{x + y + z}$

Finally, we find the partial derivative of $f$ with respect to $z$:

$\frac{\partial f}{\partial z} = \frac{\partial}{\partial z}(xz \ln(x + y + z))$

Using the product rule, we have

$\frac{\partial f}{\partial z} = x \ln(x + y + z) + xz \cdot \frac{1}{x + y + z} \cdot 1 = x \ln(x + y + z) + \frac{xz}{x + y + z}$

Therefore, the gradient is

$\nabla f = \left(z \ln(x + y + z) + \frac{xz}{x + y + z}, \frac{xz}{x + y + z}, x \ln(x + y + z) + \frac{xz}{x + y + z}\right)$.

3. Final Answer

$\nabla f = \left(z \ln(x + y + z) + \frac{xz}{x + y + z}, \frac{xz}{x + y + z}, x \ln(x + y + z) + \frac{xz}{x + y + z}\right)$