Step 1: Know the goal.
We want the maximum value of $y=x(\log x)^2$. We will use calculus: find where the slope is zero, then check which point gives a maximum.
Step 2: Differentiate using the product rule.
Treat $x$ and $(\log x)^2$ as two factors: \[ y'=(\log x)^2+x\cdot2\log x\cdot\frac{1}{x}=(\log x)^2+2\log x. \]
Step 3: Set the slope to zero.
Factor out $\log x$: \[ \log x(\log x+2)=0. \] So $\log x=0$ or $\log x=-2$, giving $x=1$ or $x=e^{-2}$.
Step 4: Decide which is the maximum.
Test the sign of $y'$ around each point. Near $x=e^{-2}$ the slope changes from positive to negative, so this is a local maximum. At $x=1$ we get a minimum.
Step 5: Plug in $x=e^{-2}$.
Here $\log x=\log(e^{-2})=-2$, so $(\log x)^2=4$. Then \[ y=e^{-2}\times4=4e^{-2}. \]
Step 6: State the answer.
The maximum value is $4e^{-2}$, matching the marked option.
\[ \boxed{4e^{-2}} \]