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To find the equation of the tangent line to the curve given by \( \frac{x}{y^3} + 2y^3 = 1 \) at the point \((-1,1)\), we need to follow these steps:

1. **Implicit Differentiation**:
   Differentiate the given equation implicitly with respect to \( x \).

2. **Solve for \(\frac{dy}{dx}\)**:
   Isolate \(\frac{dy}{dx}\) to find the slope of the tangent line.

3. **Equation of the Tangent Line**:
   Use the point-slope form of the line equation to write the tangent line's equation.

### Step-by-Step Solution:

1. **Implicit Differentiation**:
   Given the curve: 
   \[
   \frac{x}{y^3} + 2y^3 = 1
   \]
   Differentiate both sides with respect to \( x \):
   \[
   \frac{d}{dx} \left( \frac{x}{y^3} \right) + \frac{d}{dx} (2y^3) = \frac{d}{dx} (1)
   \]

   Applying the quotient rule to \( \frac{x}{y^3} \):
   \[
   \frac{d}{dx} \left( \frac{x}{y^3} \right) = \frac{y^3 \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(y^3)}{(y^3)^2}
   \]
   \[
   = \frac{y^3 \cdot 1 - x \cdot 3y^2 \frac{dy}{dx}}{y^6}
   \]
   \[
   = \frac{y^3 - 3xy^2 \frac{dy}{dx}}{y^6}
   \]
   \[
   = \frac{1}{y^3} - \frac{3x}{y^4} \frac{dy}{dx}
   \]

   Differentiating \( 2y^3 \):
   \[
   \frac{d}{dx} (2y^3) = 2 \cdot 3y^2 \frac{dy}{dx} = 6y^2 \frac{dy}{dx}
   \]

   So, the differentiated equation is:
   \[
   \frac{1}{y^3} - \frac{3x}{y^4} \frac{dy}{dx} + 6y^2 \frac{dy}{dx} = 0
   \]

2. **Solve for \(\frac{dy}{dx}\)**:
   Combine terms involving \(\frac{dy}{dx}\):
   \[
   \frac{1}{y^3} + \left(6y^2 - \frac{3x}{y^4}\right) \frac{dy}{dx} = 0
   \]
   \[
   \left(6y^2 - \frac{3x}{y^4}\right) \frac{dy}{dx} = -\frac{1}{y^3}
   \]
   \[
   \frac{dy}{dx} = -\frac{\frac{1}{y^3}}{6y^2 - \frac{3x}{y^4}}
   \]
   Simplify the denominator:
   \[
   \frac{dy}{dx} = -\frac{1}{y^3} \cdot \frac{y^4}{6y^6 - 3x}
   \]
   \[
   = -\frac{y}{6y^6 - 3x}
   \]

   Substitute \( y = 1 \) and \( x = -1 \) into the derivative:
   \[
   \left.\frac{dy}{dx}\right|_{(-1, 1)} = -\frac{1}{6(1)^6 - 3(-1)}
   \]
   \[
   = -\frac{1}{6 + 3} = -\frac{1}{9}
   \]

3. **Equation of the Tangent Line**:
   The slope of the tangent line at \((-1, 1)\) is \(-\frac{1}{9}\). Using the point-slope form \( y - y_1 = m(x - x_1) \):
   \[
   y - 1 = -\frac{1}{9}(x + 1)
   \]
   \[
   y - 1 = -\frac{1}{9}x - \frac{1}{9}
   \]
   \[
   y = -\frac{1}{9}x + \frac{8}{9}
   \]

So, the equation of the tangent line at the point \((-1, 1)\) is:
\[
y = -\frac{1}{9}x + \frac{8}{9}
\]

Would you like more details or have any questions?

### Related Questions:

1. How do you apply implicit differentiation to more complex functions involving trigonometric terms?
2. What is the geometric interpretation of the derivative at a point on a curve?
3. How can you find the normal line to a curve at a given point?
4. What are the steps to verify if a point lies on a given curve?
5. How do you compute higher-order derivatives for implicitly defined functions?

### Tip:
When using implicit differentiation, always remember to treat \( y \) as a function of \( x \) and apply the chain rule accordingly.

Consider the curve given by the equation \( \frac{x}{y^3} + 2y^3 = 1 \). Find the equation of the tangent in the point \((-1,1)\).

This solution provides a detailed step-by-step approach to finding the equation of the tangent line to the curve \( \frac{x}{y^3} + 2y^3 = 1 \) at the point \((-1,1)\). It covers key calculus concepts such as implicit differentiation, the quotient rule, and the point-slope form of a line. Suitable for university-level calculus students.

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