To find the second derivative of the implicit equation \( yx^2 + xy^2 = 1 \), let's go step-by-step using implicit differentiation.

### Step 1: Differentiate implicitly with respect to \(x\).
Given the equation:
\[
yx^2 + xy^2 = 1
\]

Differentiate both sides with respect to \(x\), remembering to apply the product rule where necessary.

For the first term, \( yx^2 \), use the product rule:
\[
\frac{d}{dx}(yx^2) = y \cdot \frac{d}{dx}(x^2) + x^2 \cdot \frac{dy}{dx} = 2yx + x^2 \frac{dy}{dx}
\]

For the second term, \( xy^2 \), again apply the product rule:
\[
\frac{d}{dx}(xy^2) = x \cdot \frac{d}{dx}(y^2) + y^2 \cdot \frac{d}{dx}(x) = x \cdot 2y \frac{dy}{dx} + y^2
\]

Now differentiate the right-hand side:
\[
\frac{d}{dx}(1) = 0
\]

Putting it all together, we get:
\[
2yx + x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + y^2 = 0
\]

### Step 2: Solve for \( \frac{dy}{dx} \).
Now, collect all terms involving \( \frac{dy}{dx} \):

\[
x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} = - (2yx + y^2)
\]

Factor out \( \frac{dy}{dx} \):

\[
(x^2 + 2xy) \frac{dy}{dx} = -(2yx + y^2)
\]

Solve for \( \frac{dy}{dx} \):

\[
\frac{dy}{dx} = \frac{-(2yx + y^2)}{x^2 + 2xy}
\]

### Step 3: Differentiate again to find the second derivative \( \frac{d^2y}{dx^2} \).
Now we need to differentiate \( \frac{dy}{dx} \) with respect to \(x\), which requires using the quotient rule. The quotient rule is given by:
\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
\]
where \( u = -(2yx + y^2) \) and \( v = x^2 + 2xy \).

1. Differentiate \( u = -(2yx + y^2) \):
   \[
   \frac{du}{dx} = - \left( 2x \frac{dy}{dx} + 2y + 2y \frac{dy}{dx} \right)
   \]

2. Differentiate \( v = x^2 + 2xy \):
   \[
   \frac{dv}{dx} = 2x + 2x \frac{dy}{dx} + 2y
   \]

Now apply the quotient rule:

\[
\frac{d^2y}{dx^2} = \frac{(x^2 + 2xy) \left( - \left( 2x \frac{dy}{dx} + 2y + 2y \frac{dy}{dx} \right) \right) - (-(2yx + y^2))(2x + 2x \frac{dy}{dx} + 2y)}{(x^2 + 2xy)^2}
\]

This is the expression for the second derivative \( \frac{d^2y}{dx^2} \).

Would you like to further simplify this or get more details on specific parts of the solution?

### 5 Relative Questions:
1. How is implicit differentiation applied in solving for higher-order derivatives?
2. What are some common errors when using the product rule in implicit differentiation?
3. How does the quotient rule apply to the second derivative calculation?
4. Why is it necessary to factor terms involving \( \frac{dy}{dx} \) when finding the first derivative?
5. Can the second derivative tell us something about the concavity of the curve defined by \( yx^2 + xy^2 = 1 \)?

**Tip**: When using implicit differentiation, always apply the product rule carefully, as multiple variables are involved.

Explore how to find the second derivative of the implicit equation yx^2 + xy^2 = 1 step-by-step. Learn the application of implicit differentiation, product rule, and quotient rule in solving this calculus problem. Ideal for students studying calculus at the undergraduate level.

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