Since $\lim_{x\to +\infty}\arctan x=\pi/2$ (resp. $\lim_{x\to -\infty}\arctan x=-\pi/2$), it is obvious the oblique asymptotes for the representative curve of this function are
$$y=x+\frac\pi2, \quad(\text{resp.} \enspace y=x-\frac\pi2).$$
General method:
Suppose a function $f$ is such that $\lim_{x\to\infty}f(x)=\infty$. One first has to compute $\displaystyle\lim_{x\to\infty}\frac{f(x)}x=\ell$. If such a limit exists, it is said that the graph of $f$ has an asymptotic direction with slope $\ell$.
- If $\ell=\infty$, we actually have a vertical parabolic branch;
- If $\ell=0$, we have a horizontal parabolic branch;
- If $\ell\neq 0,\infty$, one has to determine $\;\lim_{x\to\infty}(f(x)-\ell x)=m$.
Supposing this limit exists, if $m=\infty$, the graph has a parabolic branch in the direction with slope $\ell$. If $m<\infty$, the graph has an oblique asymptote with equation $\;y=\ell x+m$.
When possible, it is often shorter to make a change of variable $t=1/x$, and use Taylor's expansion of $f(1/t)$ near $t=0$, with two terms. In favourable cases, the expansion will look like
$$f(1/t)= \frac{\ell}t+m+o(1)=\ell x+m+o(1),$$
thus proving there is an oblique asymptote, with equation $y=\ell x+m$.