Answer: \(x=\mathrm{e}^{1/2}\) or \(x=\mathrm{e}^{3/4}\).

At a point of intersection, the two expressions for \(y\) are equal:

\(4\ln x=5-\dfrac{3}{\ln(x^2)}.\)

Since \(\ln(x^2)=2\ln x\), this becomes

\(4\ln x=5-\dfrac{3}{2\ln x}.\)

Let

\(u=\ln x.\)

Then \(u\ne0\), and the equation is

\(4u=5-\dfrac{3}{2u}.\)

Multiply by \(2u\):

\(8u^2=10u-3.\)

So

\(8u^2-10u+3=0.\)

Factorise:

\((4u-3)(2u-1)=0.\)

Thus

\(u=\frac34\quad\text{or}\quad u=\frac12.\)

Since \(u=\ln x\), the corresponding values of \(x\) are

\(x=\mathrm{e}^{3/4}\quad\text{or}\quad x=\mathrm{e}^{1/2}.\)