Since \(x>0\), \(\ln(x^4)=4\ln x\). Thus \[y=12x^3\ln x,\qquad y\prime=36x^2\ln x+12x^2=12x^2(3\ln x+1).\] As \(x\ne0\), stationarity gives \(\ln x=-1/3\), so \(x=e^{-1/3}\). Substituting, \[y=12e^{-1}\left(-\frac13\right)=-\frac4e.\] The stationary point is \[\left(e^{-1/3},-\frac4e\right).\]