Answer: \(y=\ln(18.75-3x^2)\), valid for \(-2.5\lt x\lt 2.5\).

Rewrite the exponential terms in a common form, or introduce a substitution, so that the equation becomes algebraic.

Let \(Y=\mathrm e^y\) and \(X=x^2\). Since the graph of \(Y\) against \(X\) has gradient \(-3\),

\(\mathrm e^y=-3x^2+c.\)

The line passes through \((4.30,5.85)\), so

\(5.85=-3(4.30)+c.\)

Thus

\(c=18.75.\)

Therefore

\(\mathrm e^y=18.75-3x^2,\)

and hence

\(y=\ln(18.75-3x^2).\)

For \(y\) to exist,

\(18.75-3x^2\gt 0.\)

So

\(x^2\lt 6.25.\)

Therefore

\(-2.5\lt x\lt 2.5.\)

This gives the required answer: \(y=\ln(18.75-3x^2)\), valid for \(-2.5\lt x\lt 2.5\).