Answer: \(x=\frac12\) is a maximum; \(x=3\) is a minimum.

The key point is to differentiate first, then substitute the required value or solve the resulting equation.

Differentiate

\(y=3\ln x+x^2-7x.\)

This gives

\(\frac{dy}{dx}=\frac3x+2x-7.\)

At stationary points, \(\frac{dy}{dx}=0\), so

\(\frac3x+2x-7=0.\)

Multiply by \(x\), noting that \(x\gt 0\):

\(3+2x^2-7x=0.\)

So

\(2x^2-7x+3=0.\)

Factorise:

\((2x-1)(x-3)=0.\)

Hence

\(x=\frac12\quad\text{or}\quad x=3.\)

To determine the nature, differentiate again:

\(\frac{d^2y}{dx^2}=-\frac3{x^2}+2.\)

At \(x=\frac12\),

So \(x=\frac12\) gives a maximum.

At \(x=3\),

\(\frac{d^2y}{dx^2}=-\frac3{9}+2=\frac53\gt 0.\)

So \(x=3\) gives a minimum.