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To solve the given problem, we are tasked with finding the second derivative of \( y \) with respect to \( x \) for the equation:

\[
-2x^4 + 7y^4 = 10.
\]

### Step 1: Differentiate implicitly with respect to \( x \).

We first differentiate both sides of the equation with respect to \( x \), applying implicit differentiation:

\[
\frac{d}{dx}(-2x^4 + 7y^4) = \frac{d}{dx}(10).
\]

This results in:

\[
\frac{d}{dx}(-2x^4) + \frac{d}{dx}(7y^4) = 0.
\]

Differentiating \( -2x^4 \) directly gives:

\[
\frac{d}{dx}(-2x^4) = -8x^3.
\]

For \( 7y^4 \), we use the chain rule. Differentiating with respect to \( x \) gives:

\[
\frac{d}{dx}(7y^4) = 7 \cdot 4y^3 \cdot \frac{dy}{dx} = 28y^3 \frac{dy}{dx}.
\]

Thus, the first derivative equation becomes:

\[
-8x^3 + 28y^3 \frac{dy}{dx} = 0.
\]

### Step 2: Solve for \( \frac{dy}{dx} \).

To isolate \( \frac{dy}{dx} \), rearrange the equation:

\[
28y^3 \frac{dy}{dx} = 8x^3.
\]

Divide both sides by \( 28y^3 \):

\[
\frac{dy}{dx} = \frac{8x^3}{28y^3} = \frac{2x^3}{7y^3}.
\]

### Step 3: Differentiate again to find \( \frac{d^2y}{dx^2} \).

Now, differentiate \( \frac{dy}{dx} = \frac{2x^3}{7y^3} \) with respect to \( x \) to find the second derivative. Apply the quotient rule, which states:

\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2},
\]

where \( u = 2x^3 \) and \( v = 7y^3 \).

First, compute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

\[
\frac{du}{dx} = 6x^2,
\]
\[
\frac{dv}{dx} = 7 \cdot 3y^2 \cdot \frac{dy}{dx} = 21y^2 \frac{dy}{dx}.
\]

Now, applying the quotient rule:

\[
\frac{d^2y}{dx^2} = \frac{7y^3 \cdot 6x^2 - 2x^3 \cdot 21y^2 \frac{dy}{dx}}{(7y^3)^2}.
\]

Substitute \( \frac{dy}{dx} = \frac{2x^3}{7y^3} \):

\[
\frac{d^2y}{dx^2} = \frac{42x^2 y^3 - 42x^6 / y}{49y^6}.
\]

Simplify the numerator:

\[
\frac{d^2y}{dx^2} = \frac{42x^2 y^3 - 42x^6 / y}{49y^6}.
\]

Step-by-step guide on finding the second derivative of y with respect to x for the equation -2x^4 + 7y^4 = 10 using implicit differentiation, the chain rule, and the quotient rule. Suitable for undergraduate calculus students.

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