1. **State the problem:** We need to find the partial derivative $\frac{\partial z}{\partial y}$ from the implicit equation $$x^2 - 3yz^2 + xyz - 2 = 0.$$\n\n2. **Recall the formula:** For implicit differentiation, treat $z$ as a function of $y$ (and $x$ as constant), then differentiate both sides with respect to $y$. Use the product rule and chain rule where needed.\n\n3. **Differentiate each term:**\n- $\frac{\partial}{\partial y}(x^2) = 0$ since $x$ is constant.\n- $\frac{\partial}{\partial y}(-3yz^2) = -3\left(z^2 + y \cdot 2z \frac{\partial z}{\partial y}\right) = -3z^2 - 6yz \frac{\partial z}{\partial y}$.\n- $\frac{\partial}{\partial y}(xyz) = x\left(z + y \frac{\partial z}{\partial y}\right) = xz + xy \frac{\partial z}{\partial y}$.\n- $\frac{\partial}{\partial y}(-2) = 0$.\n\n4. **Write the differentiated equation:**\n$$0 - 3z^2 - 6yz \frac{\partial z}{\partial y} + xz + xy \frac{\partial z}{\partial y} = 0.$$\n\n5. **Group terms with $\frac{\partial z}{\partial y}$:**\n$$(-6yz + xy) \frac{\partial z}{\partial y} = 3z^2 - xz.$$\n\n6. **Solve for $\frac{\partial z}{\partial y}$:**\n$$\frac{\partial z}{\partial y} = \frac{3z^2 - xz}{-6yz + xy} = \frac{3z^2 - xz}{xy - 6yz}.$$\n\n**Final answer:**\n$$\boxed{\frac{\partial z}{\partial y} = \frac{3z^2 - xz}{xy - 6yz}}.$$