Answer
Concave upward on
$$
(-\infty, -\frac{1}{\sqrt {2\ln 5}} ), \quad (\frac{1}{\sqrt {2\ln 5}} , \infty) ,
$$
concave downward on
$$
(-\frac{1}{\sqrt {2\ln 5}} , \frac{1}{\sqrt {2\ln 5}} ) ,
$$
inflection point at
$$
(-\frac{1}{\sqrt {2\ln 5}} ,e^{-(\frac{1}{2})} ), \quad (\frac{1}{\sqrt {2\ln 5}} ,e^{-(\frac{1}{2})} ).
$$
Work Step by Step
$$
f(x)=5^{-x^{2}}
$$
The first derivative is
$$
\begin{aligned}
f^{\prime}(x) &=(\ln5)5^{-x^{2}}. (-2x) \\
&=(-2x \ln5)5^{-x^{2}}
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned} f^{\prime \prime}(x) &=-2 \ln 5 \cdot 5^{-x^{2}}+(-2 \ln 5) x \cdot(\ln 5)\left(5^{-x^{2}}\right)(-2 x) \\
&=(2 \ln 5) 5^{-x^{2}}\left[2 \ln 5\left(x^{2}\right)-1\right] \end{aligned}
$$
A function $f$ is concave upward on an interval if $f^{\prime\prime}(x) \gt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) &=(2 \ln 5) 5^{-x^{2}}\left[2 \ln 5\left(x^{2}\right)-1\right] \gt 0 \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
\left[2 \ln 5\left(x^{2}\right)-1\right] & \gt 0 \\
2 \ln 5\left(x^{2}\right) & \gt 1 \\
\left(x^{2}\right) & \gt \frac{1}{2\ln 5} \\
\left(x^{2}\right) & \gt \frac{1}{2\ln 5} \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x \lt -\frac{1}{\sqrt {2\ln 5}} \quad \text {and} & \quad x \gt \frac{1}{\sqrt {2\ln 5}} \\
\end{aligned}
$$
so $f(x)$ is concave upward on $(-\infty, -\frac{1}{\sqrt {2\ln 5}} ), \quad (\frac{1}{\sqrt {2\ln 5}} , \infty)$.
Also function $f$ is concave downward on an interval if $f^{\prime\prime}(x) \lt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) &=(2 \ln 5) 5^{-x^{2}}\left[2 \ln 5\left(x^{2}\right)-1\right] \lt 0 \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
\left[2 \ln 5\left(x^{2}\right)-1\right] & \lt 0 \\
2 \ln 5\left(x^{2}\right) & \lt 1 \\
\left(x^{2}\right) & \lt \frac{1}{2\ln 5} \\
\left(x^{2}\right) & \lt \frac{1}{2\ln 5} \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
-\frac{1}{\sqrt {2\ln 5}} \lt x \lt \frac{1}{\sqrt {2\ln 5}} \\
\end{aligned}
$$
so $f(x)$ is concave downward on $(-\frac{1}{\sqrt {2\ln 5}} , \frac{1}{\sqrt {2\ln 5}} ) $ .
Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $ -\frac{1}{\sqrt {2\ln 5}} $, the graph changes from concave upward to concave downward at that point, so $(-\frac{1}{\sqrt {2\ln 5}} , f(-\frac{1}{\sqrt {2\ln 5}} ))$ is inflection point.
Also, the sign of $f^{\prime\prime}(x)$ changes from negative to positive at $\frac{1}{\sqrt {2\ln 5}} $, the graph changes from concave downward to concave upward at that point, so $(\frac{1}{\sqrt {2\ln 5}} , f(\frac{1}{\sqrt {2\ln 5}} ))$ is inflection point.
Now ,we find $f(-\frac{1}{\sqrt {2\ln 5}} ), f(\frac{1}{\sqrt {2\ln 5}} )$as follows:
$$
\begin{aligned}
f(-\frac{1}{\sqrt {2\ln 5}} ) &=5^{-(-\frac{1}{\sqrt {2\ln 5}})^{2}} \\
&=5^{-(\frac{1}{2\ln 5})} \\
&=e^{-(\frac{1}{2})}
\end{aligned}
$$
and
$$
\begin{aligned}
f(\frac{1}{\sqrt {2\ln 5}} ) &=5^{-(\frac{1}{\sqrt {2\ln 5}})^{2}} \\
&=5^{-(\frac{1}{2\ln 5})} \\
&=e^{-(\frac{1}{2})}
\end{aligned}
$$
Finally, we have inflection points at
$$
(-\frac{1}{\sqrt {2\ln 5}} ,e^{-(\frac{1}{2})} ), \quad (\frac{1}{\sqrt {2\ln 5}} ,e^{-(\frac{1}{2})} ).
$$
