Solution. Let h(x) = \ln(x^2), f(u) = \ln(u) and g(x) = x^2. We will apply that chain rule:
\begin{align*}
h'(x) = f'(g(x))g'(x).
\end{align*}\begin{align*}
f'(u) = \frac{1}{u} \quad \text{and} \quad g'(x) = 2x.
\end{align*}\begin{align*}
h'(x) &= f'(g(x))g'(x) \\
&= \frac{1}{x^2} \cdot 2x \\
&= \frac{2}{x}.
\end{align*}