Answer
$$
f(x) =\frac{x^{2}}{\ln x}
$$
A relative minimum occurs at $x=1.648,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(1.648) &=\frac{(1.648)^{2}}{\ln (1.648)}\\
&=5.436 \\
\end{aligned}
$$
Work Step by Step
$$
f(x) =\frac{x^{2}}{\ln x}
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=\frac{(\ln x) 2 x-x^{2}\left(\frac{1}{x}\right)}{(\ln x)^{2}} \\ &=\frac{2 x \ln x-x}{(\ln x)^{2}} \\
&=\frac{x(2 \ln x-1)}{(\ln x)^{2}}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime} =0 $. Now, solve the equation $f^{\prime} =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =\frac{x(2 \ln x-1)}{(\ln x)^{2}} &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x(2 \ln x-1)&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x=0 & \quad \quad or \quad \quad 2 \ln x-1=0 \\
x=0 & \quad \quad or \quad \quad 2 \ln x=1 \\
x=0 & \quad \quad or \quad \quad \ln x=\frac{1}{2} \\
x=0 & \quad \quad or \quad \quad x= e^{\frac{1}{2} } \approx 1.648\\
\end{aligned}
$$
Next, we find any values of $x$ where $f^{\prime}(x) $ fails to exist. This occurs whenever the denominator of $f^{\prime}(x) $ is $0$, here we observe that the denominator $ \ln x$ equal to 0, when $x=1$. But , $f(x) $ is not defined also at $x=1 $, since the domain $f(x)$ is $(0, \infty) .$ So,
The only critical number is : $x =e^{\frac{1}{2} } \approx 1.648 $.
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. Now, check the sign of $f^{\prime}(x)$. in the intervals
$$
(-\infty, 1.648 ), \quad\quad (1.648 ,\infty ) .
$$
(1)
Test a number in the interval $(-\infty, 1.648 )$ say $1.5$:
$$
\begin{aligned}
f^{\prime}(1.5) &=\frac{(1.5)(2 \ln (1.5)-1)}{(\ln (1.5))^{2}}\\
&=-1.725 \\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty ,1.648 )$.
(2)
Test a number in the interval $(1.648, \infty )$ say $2$:
$$
\begin{aligned}
f^{\prime}(2) &=\frac{(2)(2 \ln (2)-1)}{(\ln (2))^{2}}\\
&=1.608 \\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(1.648, \infty )$.
From (1), (2) we find that:
(*)
The function $f$ is decreasing on interval $ ( -\infty ,1.648) , $ and it is increasing on interval $ (1.648 ,\infty ) . $ So, a relative minimum occurs at $x=1.648,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(1.648) &=\frac{(1.648)^{2}}{\ln (1.648)}\\
&=5.436 \\
\end{aligned}
$$
