Answer: \(\frac{\mathrm dy}{\mathrm dx}=\frac{x^2+6x+3\lambda+6\lambda^2}{(x+3)^2}.\)

If the curve has two stationary points, then \(-\frac32\lt \lambda\lt 1\).

The asymptotes are \(x=-3\) and \(y=x+\lambda-3\).

Use the quotient rule for

\(y=\frac{x^2+\lambda x-6\lambda^2}{x+3}.\)

Then

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

Expanding the numerator gives

\((2x+\lambda)(x+3)-(x^2+\lambda x-6\lambda^2)=x^2+6x+3\lambda+6\lambda^2.\)

So

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

Stationary points occur when the numerator is zero:

\(x^2+6x+3\lambda+6\lambda^2=0.\)

For two distinct stationary points, the discriminant must be positive:

\(6^2-4(3\lambda+6\lambda^2)\gt 0.\)

Hence

\(36-12\lambda-24\lambda^2\gt 0,\)

\(3- \lambda-2\lambda^2\gt 0,\)

\(2\lambda^2+\lambda-3\lt 0.\)

Since

\(2\lambda^2+\lambda-3=(2\lambda+3)(\lambda-1),\)

we get

\(-\frac32\lt \lambda\lt 1.\)

The vertical asymptote is \(x=-3\). To find the oblique asymptote, divide the numerator by \(x+3\):

\(\frac{x^2+\lambda x-6\lambda^2}{x+3}=x+\lambda-3+\frac{9-3\lambda-6\lambda^2}{x+3}.\)

Therefore the oblique asymptote is

\(y=x+\lambda-3.\)

For \(0\lt \lambda\lt 1\), the curve has two stationary points and two branches separated by \(x=-3\), approaching the oblique asymptote \(y=x+\lambda-3\). For \(\lambda\gt 3\), the discriminant is negative, so there are no stationary points; the branches remain monotonic on each side of the vertical asymptote.