= Solution
The real <logarithm> restricts the domain to $x>0$. Differentiating gives
$$
f'(x)=2x\ln(3x)+x=x\big[2\ln(3x)+1\big].
$$
Since $x$ cannot vanish in the domain, a <stationary point> requires $\ln(3x)=-1/2$. The <logarithm> is strictly increasing, so this equation has exactly one solution:
$$
\boxed{x_* =\frac{e^{-1/2}}3=\frac1{3\sqrt e},\qquad
f(x_*)=-1-\frac1{18e}.}
$$
The <second derivative> is $f''(x)=2\ln(3x)+3$, giving $f''(x_*)=2>0$. In fact $f'$ is negative below $x_*$ and positive above it, so the point is the unique global <minimum> on $x>0$. The boundary value $x=0$ is not an additional <stationary point>, since the original function is not defined there.
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