We are tasked with finding $\frac{\partial z}{\partial x}$ at the point $(1, 1, 1)$, given the implicit equation:

$xy + z^3x - 2yz = 0$

We will use implicit differentiation to solve this problem. Here's the step-by-step process:

Step 1: Differentiate implicitly with respect to $x$

• Remember, $z$ is a function of both $x$ and $y$, so when differentiating any term involving $z$, we must apply the chain rule.

Differentiate each term:

1. $\frac{d}{dx}(xy) = y + x \frac{\partial y}{\partial x}$ but since $y$ is treated as independent of $x$, this simplifies to $y$.

2. $\frac{d}{dx}(z^3x) = z^3 + 3z^2x \frac{\partial z}{\partial x}$ (product rule applied, chain rule on $z^3$).

3. $\frac{d}{dx}(-2yz) = -2y \frac{\partial z}{\partial x}$ (since $y$ is independent of $x$, we only differentiate $z$).

Now, putting these together:

$y + z^3 + 3z^2x \frac{\partial z}{\partial x} - 2y \frac{\partial z}{\partial x} = 0$

Step 2: Solve for $\frac{\partial z}{\partial x}$

Group the terms containing $\frac{\partial z}{\partial x}$ on one side:

$3z^2x \frac{\partial z}{\partial x} - 2y \frac{\partial z}{\partial x} = -y - z^3$

Factor out $\frac{\partial z}{\partial x}$:

$\frac{\partial z}{\partial x} (3z^2x - 2y) = -y - z^3$

Finally, solve for $\frac{\partial z}{\partial x}$:

$\frac{\partial z}{\partial x} = \frac{-y - z^3}{3z^2x - 2y}$

Step 3: Substitute $x = 1, y = 1, z = 1$

Substitute the values $x = 1$, $y = 1$, and $z = 1$ into the equation:

$\frac{\partial z}{\partial x} = \frac{-(1) - 1^3}{3(1^2)(1) - 2(1)} = \frac{-1 - 1}{3 - 2} = \frac{-2}{1} = -2$

Thus, the value of $\frac{\partial z}{\partial x}$ at the point $(1, 1, 1)$ is $-2$.

Would you like any further details or clarifications?

Here are some follow-up questions for further exploration:

1. How does implicit differentiation work when the function is dependent on multiple variables?
2. What is the geometric interpretation of partial derivatives in 3D space?
3. How can we compute $\frac{\partial z}{\partial y}$ using a similar method?
4. What would change in the solution if $y$ were also dependent on $x$?
5. How does the chain rule apply in implicit differentiation?

Tip: When working with implicit differentiation, always be careful to apply the chain rule when differentiating variables that depend on the one you're differentiating with respect to.