The given problem requires showing that:

$\frac{d^2y}{dx^2} = \frac{x_1y_2 - x_2y_1}{x_1^3},$

where $x = f(t)$ and $y = g(t)$, and presumably the notations $x_1 = \frac{dx}{dt}$, $y_1 = \frac{dy}{dt}$, $x_2 = \frac{d^2x}{dt^2}$, and $y_2 = \frac{d^2y}{dt^2}$.

Solution:

Step 1: Derivative of $y$ with respect to $x$

Using the chain rule, we know: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{y_1}{x_1},$ where $y_1 = \frac{dy}{dt}$ and $x_1 = \frac{dx}{dt}$.

Step 2: Second derivative of $y$ with respect to $x$

The second derivative is found as: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{y_1}{x_1}\right).$

Step 3: Differentiate $\frac{y_1}{x_1}$ with respect to $t$

Using the quotient rule: $\frac{d}{dx}\left(\frac{y_1}{x_1}\right) = \frac{x_1 \frac{dy_1}{dt} - y_1 \frac{dx_1}{dt}}{x_1^2}.$

Now substitute $\frac{dx_1}{dt} = x_2$ and $\frac{dy_1}{dt} = y_2$: $\frac{d}{dx}\left(\frac{y_1}{x_1}\right) = \frac{x_1 y_2 - y_1 x_2}{x_1^2}.$

Step 4: Express with respect to $x$

Since $\frac{d}{dx}$ is divided by $x_1$, we have: $\frac{d^2y}{dx^2} = \frac{1}{x_1} \cdot \frac{x_1 y_2 - y_1 x_2}{x_1^2} = \frac{x_1 y_2 - y_1 x_2}{x_1^3}.$

Final Result:

$\boxed{\frac{d^2y}{dx^2} = \frac{x_1 y_2 - x_2 y_1}{x_1^3}}$

This completes the derivation.

Would you like a more detailed explanation of any step? Here are some follow-up questions:

1. What is the significance of the parametric representation $x = f(t), y = g(t)$?
2. How does the chain rule work in the context of parametric derivatives?
3. Why is the quotient rule essential in differentiating $\frac{y_1}{x_1}$?
4. Can you explain the practical applications of second derivatives in parametric equations?
5. How can the concept be extended to higher-order derivatives in parametric contexts?

Tip: When dealing with parametric equations, always compute derivatives in terms of the parameter $t$ first, and then translate them into terms of the independent variable!