Critical point via the Second Derivative Test

Using the product rule, $f'(x) = 2x\ln(x) + x^2 \cdot \frac{1}{x} = 2x\ln(x) + x = x(2\ln(x) + 1)$. Since $x > 0$ on the domain, setting $f'(x) = 0$ requires $2\ln(x) + 1 = 0$, so $\ln(x) = -\tfrac12$, giving the critical number $x = e^{-1/2}$.

Differentiating again, $f''(x) = \frac{d}{dx}\left[2x\ln(x) + x\right] = 2\ln(x) + 2 + 1 = 2\ln(x) + 3$. At $x = e^{-1/2}$, $f''(e^{-1/2}) = 2(-\tfrac12) + 3 = 2 > 0$, which is positive. By the Second Derivative Test, since $f''(x) > 0$ at this critical number, $f$ has a local minimum there.

From the previous part, $f''(x) = 2\ln(x) + 3$. Setting $f''(x) = 0$ gives $\ln(x) = -\tfrac32$, so $x = e^{-3/2}$.

For $0 < x < e^{-3/2}$, $\ln(x) < -\tfrac32$ so $f''(x) < 0$, meaning $f$ is concave down there. For $x > e^{-3/2}$, $f''(x) > 0$, so $f$ is concave up. Since concavity changes at $x = e^{-3/2}$, this point is an inflection point.