Let's analyze the function given:

$f(x, y, z) = \cos(2x) \cdot z \cdot e^{y \cdot z}$

and calculate the following partial derivatives as requested.

a) $\frac{\partial f}{\partial z}$

To find $\frac{\partial f}{\partial z}$, we differentiate $f(x, y, z)$ with respect to $z$, treating $x$ and $y$ as constants:

$f(x, y, z) = \cos(2x) \cdot z \cdot e^{y \cdot z}$

Applying the product rule, we get:

$\frac{\partial f}{\partial z} = \cos(2x) \cdot \frac{\partial}{\partial z} \left(z \cdot e^{y \cdot z}\right)$

Expanding $\frac{\partial}{\partial z} \left(z \cdot e^{y \cdot z}\right)$ with the product rule:

$\frac{\partial}{\partial z} \left(z \cdot e^{y \cdot z}\right) = e^{y \cdot z} + z \cdot y \cdot e^{y \cdot z} = e^{y \cdot z} (1 + y \cdot z)$

Thus,

$\frac{\partial f}{\partial z} = \cos(2x) \cdot e^{y \cdot z} \cdot (1 + y \cdot z)$

b) $\frac{\partial f}{\partial x}$

To find $\frac{\partial f}{\partial x}$, we differentiate $f(x, y, z)$ with respect to $x$, treating $y$ and $z$ as constants:

$f(x, y, z) = \cos(2x) \cdot z \cdot e^{y \cdot z}$

The derivative of $\cos(2x)$ with respect to $x$ is $-2 \sin(2x)$, so:

$\frac{\partial f}{\partial x} = -2 \sin(2x) \cdot z \cdot e^{y \cdot z}$

c) $\frac{\partial^2 f}{\partial x \partial y}$

To find $\frac{\partial^2 f}{\partial x \partial y}$, we first find $\frac{\partial f}{\partial y}$ and then differentiate it with respect to $x$:

$f(x, y, z) = \cos(2x) \cdot z \cdot e^{y \cdot z}$

First, we compute $\frac{\partial f}{\partial y}$:

$\frac{\partial f}{\partial y} = \cos(2x) \cdot z \cdot \frac{\partial}{\partial y} \left(e^{y \cdot z}\right) = \cos(2x) \cdot z^2 \cdot e^{y \cdot z}$

Now, we differentiate this with respect to $x$:

$\frac{\partial^2 f}{\partial x \partial y} = -2 \sin(2x) \cdot z^2 \cdot e^{y \cdot z}$

d) $\frac{\partial^2 f}{\partial y^2}$

To find $\frac{\partial^2 f}{\partial y^2}$, we first find $\frac{\partial f}{\partial y}$ and then differentiate it again with respect to $y$:

We already have:

$\frac{\partial f}{\partial y} = \cos(2x) \cdot z^2 \cdot e^{y \cdot z}$

Now, differentiate this with respect to $y$:

$\frac{\partial^2 f}{\partial y^2} = \cos(2x) \cdot z^3 \cdot e^{y \cdot z}$

Summary of the results:

1. $\frac{\partial f}{\partial z} = \cos(2x) \cdot e^{y \cdot z} \cdot (1 + y \cdot z)$
2. $\frac{\partial f}{\partial x} = -2 \sin(2x) \cdot z \cdot e^{y \cdot z}$
3. $\frac{\partial^2 f}{\partial x \partial y} = -2 \sin(2x) \cdot z^2 \cdot e^{y \cdot z}$
4. $\frac{\partial^2 f}{\partial y^2} = \cos(2x) \cdot z^3 \cdot e^{y \cdot z}$

Would you like any further explanation or details?

Here are some follow-up questions that expand on this topic:

1. How would the partial derivatives change if $f(x, y, z)$ had a different exponent in the exponential term?
2. What would be the result of $\frac{\partial^2 f}{\partial z \partial x}$?
3. Can we compute the third derivative $\frac{\partial^3 f}{\partial x \partial y \partial z}$ for this function?
4. How does the cosine function affect the partial derivatives compared to if it was a simple constant?
5. How would the answer change if $f(x, y, z) = \sin(2x) \cdot z \cdot e^{y \cdot z}$?

Tip: Always remember to apply the chain rule carefully when dealing with exponential functions and trigonometric expressions in partial derivatives.