Answer: approximate change \(=\dfrac{h(\ln3-1)}{(\ln3)^2}\).

We start with the main method. For a small change, use \(\Delta y\approx (dy/dx)\Delta x\) evaluated at the starting value.

Given

\(y=\frac{x}{\ln(3x)}.\)

Using the quotient rule,

\(\frac{\mathrm dy}{\mathrm dx} =\frac{\ln(3x)\cdot1-x\cdot\frac1x}{(\ln(3x))^2}.\)

So

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

When \(x=1\),

\(\frac{\mathrm dy}{\mathrm dx}=\frac{\ln3-1}{(\ln3)^2}.\)

For a small increase \(h\) in \(x\),

\(\Delta y\approx \frac{\mathrm dy}{\mathrm dx}\Delta x.\)

Hence the approximate change in \(y\) is

\(\frac{h(\ln3-1)}{(\ln3)^2}.\)

This completes the solution and gives the required result.