We are given the equation $3x^2z + y^3 - xyz^3 = 0$ and we need to find the partial derivative of $z$ with respect to $x$, denoted as $\frac{\partial z}{\partial x}$.

1. Problem Description

We are given the equation $3x^2z + y^3 - xyz^3 = 0$ and we need to find the partial derivative of $z$ with respect to $x$, denoted as $\frac{\partial z}{\partial x}$.

2. Solution Steps

We will use implicit differentiation to find $\frac{\partial z}{\partial x}$. We differentiate both sides of the equation with respect to $x$, treating $y$ as a constant and $z$ as a function of $x$.

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

Applying the product rule and chain rule where necessary:

$\frac{\partial}{\partial x}(3x^2z) = 3(2x \cdot z + x^2 \cdot \frac{\partial z}{\partial x}) = 6xz + 3x^2 \frac{\partial z}{\partial x}$

$\frac{\partial}{\partial x}(y^3) = 0$ (since $y$ is treated as a constant)

$\frac{\partial}{\partial x}(xyz^3) = y \frac{\partial}{\partial x}(xz^3) = y (1 \cdot z^3 + x \cdot 3z^2 \frac{\partial z}{\partial x}) = yz^3 + 3xyz^2 \frac{\partial z}{\partial x}$

So, the derivative of the entire equation with respect to $x$ is:

$6xz + 3x^2 \frac{\partial z}{\partial x} + 0 - (yz^3 + 3xyz^2 \frac{\partial z}{\partial x}) = 0$

$6xz + 3x^2 \frac{\partial z}{\partial x} - yz^3 - 3xyz^2 \frac{\partial z}{\partial x} = 0$

Now, we isolate $\frac{\partial z}{\partial x}$:

$3x^2 \frac{\partial z}{\partial x} - 3xyz^2 \frac{\partial z}{\partial x} = yz^3 - 6xz$

$\frac{\partial z}{\partial x} (3x^2 - 3xyz^2) = yz^3 - 6xz$

$\frac{\partial z}{\partial x} = \frac{yz^3 - 6xz}{3x^2 - 3xyz^2}$

We can simplify by dividing both numerator and denominator by 3:

$\frac{\partial z}{\partial x} = \frac{yz^3 - 6xz}{3x^2 - 3xyz^2} = \frac{z^3y - 6xz}{3(x^2 - xyz^2)} = \frac{yz^3 - 2(3xz)}{3x(x - yz^2)} = \frac{yz^3 - 2(3xz)}{3x(x - yz^2)} = \frac{yz^3 - 2(3xz)}{3x^2 - 3xyz^2}$

Therefore,

$\frac{\partial z}{\partial x} = \frac{yz^3 - 6xz}{3x^2 - 3xyz^2}$

3. Final Answer

$\frac{\partial z}{\partial x} = \frac{yz^3 - 6xz}{3x^2 - 3xyz^2}$