To solve this problem using implicit differentiation, we need to differentiate both sides of the given equation with respect to $x$. The given equation is:

$2x^3 + x^2 y - xy^3 = 6$

Step-by-Step Implicit Differentiation:

1. Differentiate each term with respect to $x$, treating $y$ as a function of $x$.

   • For $2x^3$: The derivative is straightforward: $\frac{d}{dx}(2x^3) = 6x^2$

   • For $x^2 y$: Apply the product rule ($uv' + vu'$), where $u = x^2$ and $v = y$: $\frac{d}{dx}(x^2 y) = 2x y + x^2 \frac{dy}{dx}$

   • For $-xy^3$: Again, use the product rule, where $u = x$ and $v = y^3$: $\frac{d}{dx}(-xy^3) = -y^3 - 3xy^2 \frac{dy}{dx}$

   • For the constant $6$: The derivative is zero: $\frac{d}{dx}(6) = 0$

2. Combine all the derivatives: $6x^2 + (2xy + x^2 \frac{dy}{dx}) - (y^3 + 3xy^2 \frac{dy}{dx}) = 0$

3. Collect the terms involving $\frac{dy}{dx}$ and the terms without it: $6x^2 + 2xy - y^3 + (x^2 - 3xy^2)\frac{dy}{dx} = 0$

4. Solve for $\frac{dy}{dx}$: $(x^2 - 3xy^2)\frac{dy}{dx} = y^3 - 6x^2 - 2xy$ $\frac{dy}{dx} = \frac{y^3 - 6x^2 - 2xy}{x^2 - 3xy^2}$

Thus, the derivative $\frac{dy}{dx}$ is: $\boxed{\frac{dy}{dx} = \frac{y^3 - 6x^2 - 2xy}{x^2 - 3xy^2}}$

Would you like more details or have any questions about this process?

5 Related Questions:

1. What is the product rule and how is it applied in differentiation?
2. Why do we treat $y$ as a function of $x$ in implicit differentiation?
3. How would this process change if there were higher powers of $y$?
4. Can implicit differentiation be used for equations involving trigonometric functions?
5. How can you verify your result after performing implicit differentiation?

Tip:

Always apply the product and chain rules carefully when differentiating implicitly, especially when differentiating terms involving both $x$ and $y$.