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To find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\), we need to differentiate the given equation implicitly with respect to \(x\).

### Step 1: Given Equation
The equation provided is:
\[
x^3 \cdot y + 6x \cdot y^5 = 2
\]

### Step 2: Differentiate Implicitly with Respect to \(x\)
Using the product rule and chain rule for each term on the left side:

1. Differentiate \(x^3 \cdot y\):
   \[
   \frac{d}{dx}(x^3 \cdot y) = x^3 \cdot \frac{dy}{dx} + 3x^2 \cdot y
   \]

2. Differentiate \(6x \cdot y^5\):
   \[
   \frac{d}{dx}(6x \cdot y^5) = 6 \cdot y^5 + 6x \cdot 5y^4 \cdot \frac{dy}{dx} = 6y^5 + 30x \cdot y^4 \cdot \frac{dy}{dx}
   \]

On the right side, since \(2\) is constant, its derivative is \(0\).

### Step 3: Combine Results
After differentiating both sides, we get:
\[
x^3 \cdot \frac{dy}{dx} + 3x^2 \cdot y + 6y^5 + 30x \cdot y^4 \cdot \frac{dy}{dx} = 0
\]

### Step 4: Solve for \(\frac{dy}{dx}\)
Collect terms involving \(\frac{dy}{dx}\) on one side:
\[
\left(x^3 + 30x \cdot y^4\right) \frac{dy}{dx} = -3x^2 \cdot y - 6y^5
\]

Now, divide by \((x^3 + 30x \cdot y^4)\) to isolate \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{-3x^2 \cdot y - 6y^5}{x^3 + 30x \cdot y^4}
\]

### Final Answer
\[
\frac{dy}{dx} = \frac{-3x^2 y - 6y^5}{x^3 + 30x y^4}
\]

Let me know if you would like further details or have any questions!

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### Related Questions:
1. How do you apply the product rule in implicit differentiation?
2. What is implicit differentiation, and when is it used?
3. How does the chain rule apply in differentiating terms with \(y\) in terms of \(x\)?
4. Can \(\frac{dy}{dx}\) have real solutions for specific values of \(x\) and \(y\) in this equation?
5. How would you solve for \(\frac{dx}{dy}\) instead?

### Tip:
When differentiating terms like \(x^n \cdot y\), remember to use the product rule to correctly account for both \(x\) and \(y\) as variable functions of \(x\).

Suppose x^3 * y + 6 * x * y^5 = 2. Find dy/dx in terms of x and y.

Learn how to find dy/dx in terms of x and y for the equation x^3 * y + 6 * x * y^5 = 2 using implicit differentiation. This solution applies the product rule and chain rule, suitable for calculus students.

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