We will be trying to find the derivative of $\ln(x^2)$. Recall the formula for the Chain Rule, which states that
\[\begin{align*} \dfrac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x). \end{align*}\]
Using this formula, we let $f(x) = \ln(x)$ and $g(x) = x^2$ so that $f(g(x)) = \ln(g(x)) = \ln(x^2)$.
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:
\[\begin{align*} f(x) = \ln(x) &\longrightarrow f'(x) = \dfrac1x,\\ g(x) = x^2 &\longrightarrow g'(x) = 2x. \end{align*}\]
Plugging these values into the Chain Rule formula, we find:
\[\begin{align*} \dfrac{d}{dx}(\ln(x^2)) = \dfrac{1}{g(x)} \cdot 2x \dfrac{d}{dx}(\ln(x^2)) = \dfrac{1}{x^2} \cdot 2x \dfrac{d}{dx}(\ln(x^2)) = \boxed{\dfrac{2}{x}}. \end{align*}\]
