(a) Substitution of \(y=e^{-1}\) gives \[-(x^2-3)+6x=14\Rightarrow x^2-6x+11=0.\] The discriminant is \(36-44=-8<0\), so there is no real \(x\), and no such point.

(b) Differentiate implicitly: \[2x\ln y+\frac{x^2-3}{y}\frac{dy}{dx}+6=0.\] Hence \[\frac{dy}{dx}=-\frac{y(2x\ln y+6)}{x^2-3}.\] At \((2,e^2)\), the gradient is \(-14e^2\). Therefore \[y-e^2=-14e^2(x-2)\Rightarrow y=-14e^2x+29e^2.\]