Answer is: option2

\( -\frac{1}{2e} \)

Solution:

Differentiate \( f(x) \) using the product rule.

\[ f(x) = x^2 \ln x \]

Let \( u = x^2 \) and \( v = \ln x \). Then,

\[ f'(x) = \frac{d}{dx} (x^2 \ln x) \]

Using the product rule:

\[ f'(x) = x^2 \cdot \frac{1}{x} + 2x \ln x \]

\[ = x + 2x \ln x \]

\[ = x(1 + 2 \ln x) \]

Set \( f'(x) = 0 \):

\[ x(1 + 2 \ln x) = 0 \]

Since \( x \neq 0 \), we solve:

\[ 1 + 2 \ln x = 0 \]

\[ 2 \ln x = -1 \]

\[ \ln x = -\frac{1}{2} \]

\[ x = e^{-1/2} = \frac{1}{\sqrt{e}} \]

Now, differentiate \( f'(x) = x(1 + 2 \ln x) \):

\[ f''(x) = 1 + 2 \ln x + 2 \]

\[ = 3 + 2 \ln x \]

Substituting \( x = \frac{1}{\sqrt{e}} \):

\[ f''\left( \frac{1}{\sqrt{e}} \right) = 3 + 2 \left( -\frac{1}{2} \right) \]

\[ = 3 - 1 = 2 > 0 \]

Since \( f''(x) > 0 \), \( x = \frac{1}{\sqrt{e}} \) is a local minimum.

Now, evaluate \( f(x) \) at the critical point:

\[ f\left( \frac{1}{\sqrt{e}} \right) = \left( \frac{1}{\sqrt{e}} \right)^2 \ln \left( \frac{1}{\sqrt{e}} \right) \]

\[ = \frac{1}{e} \times \left( -\frac{1}{2} \right) \]

\[ = -\frac{1}{2e} \]

Final Answer: \[ = -\frac{1}{2e} \]