To determine the interval on which a twice-differentiable function is convex, we need to find where its second derivative is greater than or equal to zero ($f''(x) \geq 0$).
1. Find the first derivative, $f'(x)$:
$$f(x) = x^2 \ln x$$
$$\Rightarrow f'(x) = (2x)(\ln x) + (x^2)\left(\frac{1}{x}\right)$$
$$\Rightarrow f'(x) = 2x \ln x + x$$
2. Find the second derivative, $f''(x)$:
$$f''(x) = \frac{d}{dx}(2x \ln x) + \frac{d}{dx}(x)$$
$$\Rightarrow f''(x) = \left[ (2)(\ln x) + (2x)\left(\frac{1}{x}\right) \right] + 1$$
$$\Rightarrow f''(x) = 2 \ln x + 2 + 1$$
$$\Rightarrow f''(x) = 2 \ln x + 3$$
3. Solve the inequality for convexity:
For the function to be convex, we set $f''(x) \geq 0$:
$$2 \ln x + 3 \geq 0$$
$$\Rightarrow 2 \ln x \geq -3$$
$$\Rightarrow \ln x \geq -\frac{3}{2}$$
To solve for $x$, we take the exponential of both sides. Since $e^y$ is a strictly increasing function, the direction of the inequality remains the same:
$$\Rightarrow e^{\ln x} \geq e^{-3/2}$$
$$\Rightarrow x \geq e^{-3/2}$$
Since the original domain is $x > 0$, the largest interval where the function is convex is $\boxed{[e^{-3/2}, \infty)}$.
This matches option B.
