Answer
Concave upward on
$$
(-1, 1 ) ,
$$
concave downward on
$$
(-\infty, -1 ) , \quad (1, \infty )
$$
inflection points at
$$
(-1, \ln 2 ), \quad (1, \ln2 ).
$$
Work Step by Step
$$
f(x)=\ln(x^{2}+1)
$$
The first derivative is
$$
\begin{aligned}
f^{\prime}(x) &=\frac{2x}{x^{2}+1} \\
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x) &=\frac{2(x^{2}+1)-2x (2x)}{(x^{2}+1)^{2}} \\
&=\frac{2x^{2}+2-4x^{2}}{(x^{2}+1)^{2}}\\
&=\frac{2-2x^{2}}{(x^{2}+1)^{2}}
\end{aligned}
$$
A function $f$ is concave upward on an interval if $f^{\prime\prime}(x) \gt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) &=\frac{2-2x^{2}}{(x^{2}+1)^{2}} \gt 0 \\
\Rightarrow \quad\quad\quad\\
2-2x^{2} & \gt 0\\
-2x^{2} & \gt -2 \\
x^{2} & \lt 1 \\
-1 \lt x & \lt 1
\end{aligned}
$$
so $f(x)$ is concave upward on $(-1, 1 )$.
Also function $f$ is concave downward on an interval if $f^{\prime\prime}(x) \lt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) &=\frac{2-2x^{2}}{(x^{2}+1)^{2}} \lt 0 \\
\Rightarrow \quad\quad\quad\\
2-2x^{2} & \lt 0\\
-2x^{2} & \lt -2 \\
x^{2} & \gt 1 \\
\Rightarrow \quad\quad\quad\\
x \lt -1 \quad\quad \text {and} & \quad\quad x \gt 1
\end{aligned}
$$
so $f(x)$ is concave downward on $(-\infty , -1) , \quad (1, \infty)$.
Since the sign of $f^{\prime\prime}(x)$ changes from negative to positive at $ -1 $, the graph changes from concave downward to concave upward at that point, so $(-1 , f(-1)$ is inflection point.
Also, the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $1 $, the graph changes from concave upward to concave downward at that point, so $(1 , f(1)$ is inflection point.
Now ,we find $f( -1), f(1)$as follows:
$$
f(-1)=\ln((-1)^{2}+1)=\ln2,\\
f(1)=\ln((1)^{2}+1)=\ln2,
$$
Finally, we have inflection points at $(-1, \ln 2 ), \quad (1, \ln2 ) $.
