To solve for $\frac{d^2y}{dx^2}$ at the point $(2, -1)$ given the equation:

$-x^2 - 5y + y^3 = 0,$

we'll go through the following steps:

Step 1: Differentiate Implicitly with Respect to $x$

Since $y$ is a function of $x$, we use implicit differentiation.

1. Differentiate each term with respect to $x$: $\frac{d}{dx}(-x^2) + \frac{d}{dx}(-5y) + \frac{d}{dx}(y^3) = 0.$

2. Differentiating each term:

   • For $-x^2$: $\frac{d}{dx}(-x^2) = -2x$.
   • For $-5y$: $\frac{d}{dx}(-5y) = -5 \frac{dy}{dx} = -5y'$.
   • For $y^3$: $\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx} = 3y^2 y'$.

3. Substitute these results into the equation: $-2x - 5y' + 3y^2 y' = 0.$

4. Solve for $y'$: $-2x = y'(5 - 3y^2).$ $y' = \frac{-2x}{5 - 3y^2}.$

Step 2: Differentiate $y'$ Again to Find $y''$

Now we need to differentiate $y'$ with respect to $x$ to find $\frac{d^2y}{dx^2}$.

1. Differentiate $y' = \frac{-2x}{5 - 3y^2}$ with respect to $x$: $y'' = \frac{d}{dx} \left( \frac{-2x}{5 - 3y^2} \right).$

2. Use the quotient rule: $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$, where:

   • $u = -2x$, so $u' = -2$.
   • $v = 5 - 3y^2$, so $v' = -6y y'$.

3. Substitute into the quotient rule: $y'' = \frac{(-2)(5 - 3y^2) - (-2x)(-6y y')}{(5 - 3y^2)^2}.$

4. Simplify: $y'' = \frac{-10 + 6y^2 - 12xy y'}{(5 - 3y^2)^2}.$

Step 3: Substitute $x = 2$, $y = -1$, and $y'$

1. First, substitute $x = 2$ and $y = -1$ into $y'$: $y' = \frac{-2(2)}{5 - 3(-1)^2} = \frac{-4}{5 - 3} = \frac{-4}{2} = -2.$

2. Now substitute $x = 2$, $y = -1$, and $y' = -2$ into $y''$: $y'' = \frac{-10 + 6(-1)^2 - 12(2)(-1)(-2)}{(5 - 3(-1)^2)^2}.$

3. Simplify each part:

   • The numerator becomes: $-10 + 6(1) - 12 \cdot 2 \cdot 1 \cdot 2 = -10 + 6 - 48 = -52.$
   • The denominator becomes: $(5 - 3 \cdot 1)^2 = (5 - 3)^2 = 2^2 = 4.$

4. Thus, $y'' = \frac{-52}{4} = -13.$

Final Answer

$\frac{d^2y}{dx^2} = -13 \quad \text{at the point} \quad (2, -1).$

Would you like further details on any of these steps?

Here are some related questions for deeper understanding:

1. How do we apply implicit differentiation to other types of equations?
2. What is the purpose of the quotient rule in differentiation?
3. How do we interpret the second derivative geometrically?
4. What happens if we change the point at which we find $\frac{d^2y}{dx^2}$?
5. Can implicit differentiation be used for non-polynomial equations?

Tip: Always verify if simplification of intermediate steps is possible before calculating final values, as it can save time and reduce errors.