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Derivatives are a fundamental concept in calculus, representing the rate at which a function is changing at any given point. When we're looking at the function of a curve on a graph, finding the derivative is akin to determining the slope of the tangent line at any point on that curve. To find a derivative, we typically use rules of differentiation, like the power rule, product rule, and quotient rule.

The example provided involves finding the derivative of two separate functions, with respect to 't'. In this context, we are dealing with functions that have different independent variables. The first step invokes the power rule, which tells us that the derivative of any variable raised to a power, like the cubic root of 't' in the function for 'x', will be the power multiplied by the variable to the power decreased by one. Hence, for the function 'x', we get \( dx/dt = \frac{1}{3}t^{-2/3} \). Similarly, for the linear function 'y', the change rate is constant, thus its derivative is simply \( dy/dt = -1 \) as provided in the steps.

Implicit differentiation is a technique we use when a function is not given in the standard form y=f(x), but instead the relation between y and x is given in an equation that mixes x and y together. This approach allows us to differentiate with respect to 'x' without having to solve the equation for 'y' explicitly.

When we apply implicit differentiation to the exercise, we don't have an explicit function for 'y' in terms of 'x', but we do have their relationships in terms of 't'. To find \( dy/dx \) implicitly, we first find \( dy/dt \) and \( dx/dt \) separately and then divide them, because the derivative of 'y' with respect to 'x' can be linked through 't' using the chain rule. The process is made clear in the steps solution, which intuitively tells us how fast 'y' is changing in relation to 'x' by using 't' as the intermediary.

The chain rule is an essential tool in calculus for finding the derivative of a composition of functions. It can be especially useful when a function is nested within another function, and we want to understand how changes in the input affect changes in the output overall. The formula for the chain rule is encapsulated as \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \), where 'u' is an intermediary function.

In this exercise, we consider 't' as our intermediary, allowing us to find \( dy/dx \) by computing the derivatives of 'x' and 'y' with respect to 't', and then dividing them. The provided solution perfectly demonstrates the application of the chain rule in an indirect manner, where \( dy/dx = -1 / (\frac{1}{3}t^{-2/3}) \), which simplifies to \( -3t^{2/3} \). This final expression represents how 'y' changes with respect to 'x', a nuanced understanding made possible by the intuitive and powerful chain rule.