#### Answer

1) Check the continuity of $f(x)=1-x^2$
2) Check the continuity of $f(x)=\ln x$
3) Check the continuity of $f(x)$ at $1$.

#### Work Step by Step

$1-x^2$ is continuous on $R$, so it is continuous on $(-\infty,1)$
$\ln x$ is continuous on $(0,\infty)$, so it is continuous on $(1,\infty)$
Therefore, $f(x)$ is continuous on $(-\infty,1)\cup(1,\infty)$
Now we only need to check if $f(x)$ is continuous at $1$.
By definition, $f(x)$ is continuous at $1$ if and only if $$\lim\limits_{x\to1}f(x)=f(1)$$
1) For $x\lt;1$, we have
$\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^-}(1-x^2)=1-1^2=0$
2) For $x\gt1$, we have
$\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^+}\ln x=\ln1=0$
Since $\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^+}f(x)=0$
Therefore, $\lim\limits_{x\to1}f(x)=0$
Notice that $f(1)=1-1^2=0$
Since $\lim\limits_{x\to1}f(x)=f(1)$, $f(x)$ is continuous at $1$.
Overall, $f(x)$ is continuous on $(-\infty,\infty)$.