We are asked to find the partial derivative of $z$ with respect to $x$ at the point $(0, 0, 0)$ for the implicit function:

$f(x, y, z) = x^3 + z^3 + x e^{yz} + z \cos x = 0.$

To solve this, we will apply implicit differentiation with respect to $x$, treating $z$ as a function of $x$, i.e., $z = z(x, y)$. Let's go step by step.

Step 1: Differentiate implicitly with respect to $x$

We differentiate both sides of the equation $f(x, y, z) = 0$ with respect to $x$:

$\frac{\partial}{\partial x} \left( x^3 + z^3 + x e^{yz} + z \cos x \right) = \frac{\partial}{\partial x} (0).$

On the left-hand side, we apply the chain rule, remembering that $z$ is a function of $x$, so terms involving $z$ will require us to include $\frac{\partial z}{\partial x}$.

Step 2: Differentiate each term

1. $\frac{\partial}{\partial x} \left( x^3 \right) = 3x^2$.
2. $\frac{\partial}{\partial x} \left( z^3 \right) = 3z^2 \cdot \frac{\partial z}{\partial x}$.
3. $\frac{\partial}{\partial x} \left( x e^{yz} \right)$ requires the product rule: $\frac{\partial}{\partial x} \left( x e^{yz} \right) = e^{yz} + x \cdot e^{yz} \cdot \left( y \cdot \frac{\partial z}{\partial x} \right).$
4. $\frac{\partial}{\partial x} \left( z \cos x \right)$ also requires the product rule: $\frac{\partial}{\partial x} \left( z \cos x \right) = \frac{\partial z}{\partial x} \cdot \cos x - z \cdot \sin x.$

Step 3: Combine the derivatives

Putting everything together, we get the following equation:

$3x^2 + 3z^2 \cdot \frac{\partial z}{\partial x} + e^{yz} + x \cdot e^{yz} \cdot \left( y \cdot \frac{\partial z}{\partial x} \right) + \frac{\partial z}{\partial x} \cdot \cos x - z \cdot \sin x = 0.$

Step 4: Evaluate at $(0, 0, 0)$

Now, substitute $x = 0$, $y = 0$, and $z = 0$ into the equation:

1. $3x^2 = 3(0)^2 = 0$,
2. $3z^2 \cdot \frac{\partial z}{\partial x} = 0 \cdot \frac{\partial z}{\partial x} = 0$,
3. $e^{yz} = e^{0 \cdot 0} = 1$,
4. $x \cdot e^{yz} \cdot ( y \cdot \frac{\partial z}{\partial x} ) = 0 \cdot 1 \cdot ( 0 \cdot \frac{\partial z}{\partial x} ) = 0$,
5. $\frac{\partial z}{\partial x} \cdot \cos x = \frac{\partial z}{\partial x} \cdot \cos(0) = \frac{\partial z}{\partial x} \cdot 1$,
6. $- z \cdot \sin x = - 0 \cdot \sin(0) = 0$.

Thus, the equation simplifies to:

$1 + \frac{\partial z}{\partial x} = 0.$

Step 5: Solve for $\frac{\partial z}{\partial x}$

From the simplified equation, we find:

$\frac{\partial z}{\partial x} = -1.$

Final Answer:

The value of $\frac{\partial z}{\partial x}$ at the point $(0, 0, 0)$ is $\boxed{-1}$.

Let me know if you'd like further details or have any questions.

Here are some follow-up questions:

1. How does the chain rule apply in implicit differentiation?
2. What happens when $y$ is not 0 in this problem?
3. Can we generalize this method to find $\frac{\partial z}{\partial y}$?
4. How would the solution change if the equation involved trigonometric functions of $y$?
5. Can we use this process to find second-order partial derivatives?

Tip: When differentiating implicitly, remember to treat the dependent variable (in this case, $z$) as a function of the independent variables and apply the chain rule accordingly.