In this article, we will be trying to show that:

\[\begin{align*} \dfrac{d}{dx}(\ln(x^3)) = \boxed{\dfrac3x}. \end{align*}\]

To start our proof, we will use the Power Rule for logarithms, which tells us $\ln(x^n) = n \ln (x)$. Using this rule, we get that $\ln(x^3) = 3\ln(x)$. Using the fact that $\dfrac{d}{dx} (\ln(x)) = \dfrac{1}{x}$, we have:

\[\begin{align*} \dfrac{d}{dx}(\ln(x^3)) = \dfrac{d}{dx}(3\ln(x)) =\\ 3 \cdot \dfrac{d}{dx} (\ln (x)) = 3 \cdot \dfrac{1}{x} =\\ \boxed{\dfrac{3}{x}}. \end{align*}\]

Another way we could approach this problem is using the Chain Rule, which states:

\[\begin{align*} \dfrac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x). \end{align*}\]

Where $f(x)$ and $g(x)$ are differentiable functions. Using this formula, we let $f(x) = \ln (x)$ and $g(x) = x^3$ such that $f(g(x)) = \ln(x^3)$. Both of these formulas are differentiable, meaning we can use them in the Chain Rule.

To plug these values into the Chain Rule, we need to calculate their derivatives. We can use the formula for the derivative of $\ln(x)$ and the Power Rule:

\[\begin{align*} \dfrac{d}{dx}(\ln(x)) = \dfrac1x \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \dfrac{d}{dx}(x^n)= nx^{(n-1)} \end{align*}\]

Using this, we find:

\[\begin{align*} f(x) = \ln(x) &\longrightarrow f'(x) = \dfrac1x\\ g(x) = x^3 &\longrightarrow g'(x) = 3x^2. \end{align*}\]

Plugging these values into the Chain Rule formula, we find:

\[\begin{align*} \dfrac{d}{dx}(ln(x^3)) = \dfrac{1}{g(x)} \cdot 3x^2.\\ \dfrac{d}{dx}(ln(x^3)) = \dfrac{1}{x^3} \cdot 3x^2 \\ \dfrac{d}{dx}(ln(x^3)) = \boxed{\dfrac{3}{x}}. \end{align*}\]