Answer:

1. The asymptotes are \(x=-\frac12\) and \(y=\frac12x-\frac14\).
2. The stationary points are \((0,0)\) and \((-1,-1)\).
3. The sketch has vertical asymptote \(x=-\frac12\), oblique asymptote \(y=\frac12x-\frac14\), an upper branch for \(x>-\frac12\) with minimum at \((0,0)\), and a lower branch for \(x<-\frac12\) with maximum at \((-1,-1)\).

1. The vertical asymptote occurs where the denominator is zero:

   \(2x+1=0\), so \(x=-\frac12\).

   Now divide \(x^2\) by \(2x+1\):

   \(\displaystyle \frac{x^2}{2x+1}=\frac12x-\frac14+\frac{1}{4(2x+1)}\).

   As \(|x|\to\infty\), the final term tends to \(0\), so the oblique asymptote is

   \(y=\frac12x-\frac14\).

   Therefore the asymptotes are \(x=-\frac12\) and \(y=\frac12x-\frac14\).

2. Using

   \(\displaystyle y=\frac12x-\frac14+\frac{1}{4(2x+1)}\),

   differentiate:

   \(\displaystyle \frac{dy}{dx}=\frac12-\frac{1}{2(2x+1)^2}\).

   At a stationary point, \(\frac{dy}{dx}=0\), so

   \(\displaystyle \frac12-\frac{1}{2(2x+1)^2}=0\).

   Hence \((2x+1)^2=1\), giving

   \(2x+1=\pm1\).

   So \(x=0\) or \(x=-1\).

   Substituting into \(y=\frac{x^2}{2x+1}\):

   when \(x=0\), \(y=0\); when \(x=-1\), \(y=-1\).

   Therefore the stationary points are \((0,0)\) and \((-1,-1)\).

3. For the sketch, draw the vertical asymptote \(x=-\frac12\) and the oblique asymptote \(y=\frac12x-\frac14\).

   Since

   \(\displaystyle y-\left(\frac12x-\frac14\right)=\frac{1}{4(2x+1)}\),

   the curve is below the oblique asymptote for \(x<-\frac12\), and above it for \(x>-\frac12\).

   As \(x\to-\frac12^-\), \(y\to-\infty\), and as \(x\to-\frac12^+\), \(y\to+\infty\).

   The left-hand branch approaches the oblique asymptote from below as \(x\to-\infty\), has a stationary point at \((-1,-1)\), and then tends to \(-\infty\) as it approaches the vertical asymptote.

   The right-hand branch tends to \(+\infty\) as it approaches the vertical asymptote from the right, has a stationary point at \((0,0)\), and then approaches the oblique asymptote from above as \(x\to+\infty\).