Given the implicitly defined function $y = y(x)$ by the equation

$x^3 + y^3 - 6xy = 1$

we'll calculate $y'(0)$ and $y''(0)$.

(a) Find $y'(0)$

To find $y'(x)$, we’ll implicitly differentiate both sides of the equation with respect to $x$:

1. Differentiate $x^3$: $\frac{d}{dx}(x^3) = 3x^2$

2. Differentiate $y^3$ using the chain rule: $\frac{d}{dx}(y^3) = 3y^2 \cdot y'$

3. Differentiate $-6xy$ using the product rule: $\frac{d}{dx}(-6xy) = -6(y + x y')$

Differentiating the entire equation $x^3 + y^3 - 6xy = 1$ with respect to $x$:

$3x^2 + 3y^2 y' - 6(y + x y') = 0$

Now, rearrange terms to solve for $y'$:

$3y^2 y' - 6x y' = 6y - 3x^2$

$y'(3y^2 - 6x) = 6y - 3x^2$

$y' = \frac{6y - 3x^2}{3y^2 - 6x}$

To find $y'(0)$, we substitute $x = 0$ into the equation. First, we need to determine $y(0)$.

Step 1: Find $y(0)$

Substitute $x = 0$ in the original equation:

$0^3 + y^3 - 6 \cdot 0 \cdot y = 1$

$y^3 = 1 \Rightarrow y = 1$

So, $y(0) = 1$.

Step 2: Substitute $x = 0$ and $y = 1$ into $y'$

$y'(0) = \frac{6 \cdot 1 - 3 \cdot 0^2}{3 \cdot 1^2 - 6 \cdot 0} = \frac{6}{3} = 2$

Thus,

$y'(0) = 2$

(b) Find $y''(0)$

To find $y''(0)$, we need to differentiate $y'$ with respect to $x$ and evaluate it at $x = 0$.

Given

$y' = \frac{6y - 3x^2}{3y^2 - 6x}$

Differentiating both the numerator and the denominator with respect to $x$ and applying the quotient rule will yield $y''$. This calculation involves applying implicit differentiation again to account for $y = y(x)$ in both terms.

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