What is the value of $\arctan \left(\frac xy\right) +\arctan \left(\frac yx\right)?$

Yes it's a correct method.

As an alternative note that for $x>0$

$$\arctan x + \arctan \frac1x = \frac{\pi}2$$

indeed if you set

$$y=\arctan \frac1x$$

then

$$\tan y=\frac1x$$

that is

$$x=\cot y=\tan\left(\frac{\pi}{2}-y\right)$$

therefore

$$\arctan x=\arctan\tan\left(\frac{\pi}{2}-y\right)=\frac{\pi}{2}-y=\frac{\pi}{2}-\arctan \frac1x$$