Answer
$y=\pm (x+1)$ or $y=|x+1|$
Work Step by Step
Step 1. Given the function $y=\sqrt {x^2+2x}=\sqrt {(x+1)^2-1}$, we need to find all the oblique asymptotes.
Step 2. We need to find the end behavior or the limits of the function when $x\to\pm\infty$.
Step 3. Assume $x\gt-1$, $\lim_{x\to\infty}y=\lim_{x\to\infty}\sqrt {(x+1)^2-1}=\lim_{x\to\infty}(x+1)\sqrt {1-\frac{1}{(x+1)^2}}=x+1$
Step 4. Assume $x\lt-1$, $\lim_{x\to-\infty}y=\lim_{x\to-\infty}\sqrt {(x+1)^2-1}=\lim_{x\to-\infty}(-x-1)\sqrt {1-\frac{1}{(x+1)^2}}=-(x+1)$
Step 5. Thus, the oblique asymptotes are $y=\pm (x+1)$ or $y=|x+1|$
