Answer: \(\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac14\) at \(x=0\), and \(\frac{\mathrm{d}x}{\mathrm{d}t}=-2\).

(a) The curve is

\(y=\dfrac{\mathrm e^{x^2}}{x-2}.\)

Using the quotient rule,

\(\dfrac{\mathrm dy}{\mathrm dx} =\dfrac{(x-2)(2x\mathrm e^{x^2})-\mathrm e^{x^2}}{(x-2)^2}.\)

Factorising \(\mathrm e^{x^2}\),

\(\dfrac{\mathrm dy}{\mathrm dx} =\dfrac{\mathrm e^{x^2}\left(2x(x-2)-1\right)}{(x-2)^2}.\)

When \(x=0\),

\(\dfrac{\mathrm dy}{\mathrm dx} =\dfrac{1(0-1)}{(-2)^2}=-\dfrac14.\)

(b) The rate of change of \(y\) with respect to time is related to the rate of change of \(x\) by

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

At \(x=0\), \(\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac14\). It is given that \(y\) is increasing at \(0.5\) units per second, so

\(\dfrac{\mathrm dy}{\mathrm dt}=0.5=\dfrac12.\)

Hence

\(\dfrac12=-\dfrac14\dfrac{\mathrm dx}{\mathrm dt}.\)

Multiplying by \(-4\),

\(\dfrac{\mathrm dx}{\mathrm dt}=-2.\)