[Cambridge/9231/2006]
The variables $x$ and $y$ are such that $y^3 + \left(\frac{\text{d}y}{\text{d}x}\right)^3 = x^4 + 6.$ Given that $y = -1$ when $x = 1$, find the value of $\frac{\text{d}^2y}{\text{d}x^2}$ when $x = 1$.

1. Substitute the given values into the original equation to find an equation for the first derivative.

When $x = 1$, we are given $y = -1$. Substituting these values into the equation $y^3 + \left(\frac{\text{d}y}{\text{d}x}\right)^3 = x^4 + 6$:
$(-1)^3 + \left(\frac{\text{d}y}{\text{d}x}\right)^3 = (1)^4 + 6$
$-1 + \left(\frac{\text{d}y}{\text{d}x}\right)^3 = 7$

2. Solve for the value of the first derivative at $x = 1$.

$\left(\frac{\text{d}y}{\text{d}x}\right)^3 = 8$
$\frac{\text{d}y}{\text{d}x} = 2$

3. Differentiate the original equation implicitly with respect to $x$ to introduce the second derivative.

Differentiating $y^3 + \left(\frac{\text{d}y}{\text{d}x}\right)^3 = x^4 + 6$ implicitly with respect to $x$:
$\frac{\text{d}}{\text{d}x}(y^3) + \frac{\text{d}}{\text{d}x}\left(\left(\frac{\text{d}y}{\text{d}x}\right)^3\right) = \frac{\text{d}}{\text{d}x}(x^4 + 6)$
$3y^2\frac{\text{d}y}{\text{d}x} + 3\left(\frac{\text{d}y}{\text{d}x}\right)^2\frac{\text{d}^2y}{\text{d}x^2} = 4x^3$

4. Substitute the known values of $x$, $y$, and $\frac{\text{d}y}{\text{d}x}$ into the differentiated equation to solve for $\frac{\text{d}^2y}{\text{d}x^2}$.

Substitute $x = 1$, $y = -1$, and $\frac{\text{d}y}{\text{d}x} = 2$ into the new equation:
$3(-1)^2(2) + 3(2)^2\frac{\text{d}^2y}{\text{d}x^2} = 4(1)^3$
$6 + 12\frac{\text{d}^2y}{\text{d}x^2} = 4$
$12\frac{\text{d}^2y}{\text{d}x^2} = -2$
$\frac{\text{d}^2y}{\text{d}x^2} = -\frac{1}{6}$