Answer: \(m=\dfrac14\), \(n=-\dfrac54\).

Given

\(y=mx^2+\frac{x}{2}+n.\)

Differentiate once:

\(\frac{dy}{dx}=2mx+\frac12.\)

Differentiate again:

\(\frac{d^2y}{dx^2}=2m.\)

Use the given identity

\(3\frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2-y.\)

This gives

\(6m=\left(2mx+\frac12\right)^2-\left(mx^2+\frac{x}{2}+n\right).\)

Expand the right-hand side:

\(\left(2mx+\frac12\right)^2-\left(mx^2+\frac{x}{2}+n\right) =(4m^2-m)x^2+\left(2m-\frac12\right)x+\left(\frac14-n\right).\)

This must be equal to the constant \(6m\) for all values of \(x\).

Therefore the coefficients of \(x^2\) and \(x\) are zero.

From \(4m^2-m=0\),

\(m(4m-1)=0.\)

Since \(m\ne0\),

\(m=\frac14.\)

Now compare the constant terms:

\(6m=\frac14-n.\)

Substitute \(m=\frac14\):

\(\frac32=\frac14-n.\)

Hence

\(n=-\frac54.\)

This completes the solution and gives the required result.