Answer: \(P=\left(0,\ln8-\dfrac13\right)\).

Differentiate

\(y=\ln(3x^2-4)-\frac{x^3}{6}.\)

This gives

\(\frac{dy}{dx}=\frac{6x}{3x^2-4}-\frac{x^2}{2}.\)

At \(x=2\),

\(\frac{dy}{dx}=\frac{12}{12-4}-\frac{4}{2} =\frac{12}{8}-2 =-\frac12.\)

Also, when \(x=2\),

\(y=\ln(3(2)^2-4)-\frac{2^3}{6} =\ln8-\frac43.\)

The tangent has gradient \(-\frac12\) and passes through

\(\left(2,\ln8-\frac43\right).\)

So its equation is

\(y-\left(\ln8-\frac43\right)=-\frac12(x-2).\)

At the \(y\)-axis, \(x=0\). Therefore

\(y-\left(\ln8-\frac43\right)=1.\)

Hence

\(y=\ln8-\frac13.\)

So

\(P=\left(0,\ln8-\frac13\right).\)