1. The problem is to determine the continuity of the function $f(x) = \ln(x^2)$.\n\n2. First, recall that the natural logarithm function $\ln(x)$ is defined only for $x > 0$.\n\n3. Since $f(x) = \ln(x^2)$, the argument inside the logarithm is $x^2$.\n\n4. Note that $x^2 \geq 0$ for all real $x$, but $\ln(x^2)$ requires $x^2 > 0$. So $f(x)$ is defined for all $x \neq 0$.\n\n5. Therefore, the domain of $f$ is $(-\infty, 0) \cup (0, \infty)$.\n\n6. On each of these intervals, $x^2$ is positive and continuous, and $\ln$ is continuous on $(0, \infty)$, so $f(x)$ is continuous on $(-\infty, 0)$ and $(0, \infty)$.\n\n7. At $x=0$, $f(x)$ is not defined, so $f$ is not continuous at $x=0$.\n\n\nFinal answer: $f(x) = \ln(x^2)$ is continuous for all $x \neq 0$ and discontinuous at $x=0$.