Answer: \(\frac{\mathrm dx}{\mathrm dt}=\frac{9h}{9-\ln2}\).

Use differentiation to find the gradient information needed. Stationary points occur when the derivative is zero, and tangents or normals are found from the gradient at the given point.

\(y=\frac{\ln(3x^2-1)}{x+2}\).

Using the quotient rule,

\(\frac{\mathrm dy}{\mathrm dx}=\frac{(x+2)\frac{6x}{3x^2-1}-\ln(3x^2-1)}{(x+2)^2}\).

At \(x=1\),

\(\frac{\mathrm dy}{\mathrm dx}=\frac{3\cdot3-\ln2}{9}=\frac{9-\ln2}{9}\).

Since \(\frac{\mathrm dy}{\mathrm dt}=h\),

\(\frac{\mathrm dy}{\mathrm dt}=\frac{\mathrm dy}{\mathrm dx}\frac{\mathrm dx}{\mathrm dt}\).

So

\(h=\frac{9-\ln2}{9}\frac{\mathrm dx}{\mathrm dt}\), and hence

\(\frac{\mathrm dx}{\mathrm dt}=\frac{9h}{9-\ln2}\).

This gives the required answer: \(\frac{\mathrm dx}{\mathrm dt}=\frac{9h}{9-\ln2}\).