To find the gradient of the function $f(x, y, z)$, we first need to compute the partial derivatives of $f$ with respect to each variable $x$, $y$, and $z$.

Function:

$f(x, y, z) = x^2 + y^2 - 3z^2 + z \ln x$

Step 1: Compute the partial derivatives

1. Partial derivative of $f$ with respect to $x$: $\frac{\partial f}{\partial x} = 2x + \frac{z}{x}$

2. Partial derivative of $f$ with respect to $y$: $\frac{\partial f}{\partial y} = 2y$

3. Partial derivative of $f$ with respect to $z$: $\frac{\partial f}{\partial z} = -6z + \ln x$

Step 2: Evaluate the gradient at the point $(1, 5, 3)$

Substitute $x = 1$, $y = 5$, and $z = 3$ into the partial derivatives:

1. For $\frac{\partial f}{\partial x}$ at $(1, 5, 3)$: $\frac{\partial f}{\partial x} = 2(1) + \frac{3}{1} = 2 + 3 = 5$

2. For $\frac{\partial f}{\partial y}$ at $(1, 5, 3)$: $\frac{\partial f}{\partial y} = 2(5) = 10$

3. For $\frac{\partial f}{\partial z}$ at $(1, 5, 3)$: $\frac{\partial f}{\partial z} = -6(3) + \ln(1) = -18 + 0 = -18$

Step 3: Gradient at the point $(1, 5, 3)$

The gradient of $f(x, y, z)$ at $(1, 5, 3)$ is:

$\boxed{\nabla f = (5, 10, -18)}$

Would you like more details on any specific part of the solution, or do you have any questions?

Here are 5 related questions:

1. How do we interpret the gradient geometrically?
2. What does the gradient tell us about the function at the given point?
3. How would the gradient change if the point $(x, y, z)$ were different?
4. Can you compute the directional derivative of $f$ in a given direction at this point?
5. What is the second derivative (Hessian matrix) of $f(x, y, z)$?

Tip: The gradient vector points in the direction of the steepest ascent of a function at a given point.