The problem asks us to implicitly differentiate $\frac{d^2y}{dx^2}$ from the equation:

$x^2 y - 3x = 6$

Step 1: First derivative ( $\frac{dy}{dx}$ )

Differentiate both sides of the equation implicitly with respect to $x$:

$\frac{d}{dx}(x^2 y - 3x) = \frac{d}{dx}(6)$

Using the product rule for the term $x^2 y$ and applying simple differentiation for $3x$, we get:

$2x y + x^2 \frac{dy}{dx} - 3 = 0$

Now, solve for $\frac{dy}{dx}$:

$x^2 \frac{dy}{dx} = -2x y + 3$ $\frac{dy}{dx} = \frac{-2x y + 3}{x^2}$

Step 2: Second derivative ( $\frac{d^2y}{dx^2}$ )

Now, differentiate $\frac{dy}{dx}$ implicitly again with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{-2x y + 3}{x^2} \right)$

Apply the quotient rule to differentiate the right-hand side:

$\frac{d}{dx} \left( \frac{-2x y + 3}{x^2} \right) = \frac{(x^2)(\frac{d}{dx}(-2x y + 3)) - (-2x y + 3)(\frac{d}{dx}(x^2))}{(x^2)^2}$

First, differentiate the numerator $(-2x y + 3)$: $\frac{d}{dx}(-2x y + 3) = -2 \left( y + x \frac{dy}{dx} \right)$

Now, substitute this into the formula for the quotient rule:

$\frac{d^2 y}{dx^2} = \frac{x^2 (-2 (y + x \frac{dy}{dx})) - (-2x y + 3)(2x)}{x^4}$

Simplify the expression:

$\frac{d^2 y}{dx^2} = \frac{-2x^2 (y + x \frac{dy}{dx}) + 4x^2 y - 6x}{x^4}$ $\frac{d^2 y}{dx^2} = \frac{-2x^3 \frac{dy}{dx} - 6x}{x^4}$ Finally, plug in $\frac{dy}{dx} = \frac{-2x y + 3}{x^2}$ into the expression:

$\frac{d^2 y}{dx^2} = \frac{-2x^3 \left( \frac{-2x y + 3}{x^2} \right) - 6x}{x^4}$ Simplify this expression:

$\frac{d^2 y}{dx^2} = \frac{-2x \left( -2x y + 3 \right) - 6x}{x^4}$ $\frac{d^2 y}{dx^2} = \frac{4x^2 y - 6x - 6x}{x^4}$ $\frac{d^2 y}{dx^2} = \frac{4x^2 y - 12x}{x^4}$ Finally:

$\frac{d^2 y}{dx^2} = \frac{4y - 12/x}{x^2}$

Thus, the second derivative is:

$\boxed{\frac{4y - \frac{12}{x}}{x^2}}$

Let me know if you'd like more details or have questions! Here are 5 related questions:

1. How do you apply the product rule to implicitly differentiate?
2. What is the quotient rule and how does it apply here?
3. How do you simplify expressions when differentiating?
4. How can you find the second derivative from an implicit function?
5. Why do we use implicit differentiation in this case?

Tip: When using implicit differentiation, always treat $y$ as a function of $x$ even if it doesn't look explicit.