LHS = $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) $
= $\frac{\pi}{4}+ \tan^{-1}\left(\frac{2+3}{1-2.3}\right)$
= $\frac{\pi}{4}+\tan^{-1}{5\over-5}$
= $\frac{\pi}{4}+ \tan^{-1}(-1)$
=$\frac{\pi}{4}-\tan^{-1}(1)$
=$\frac{\pi}{4}-\frac{\pi}{4}$
= 0
Where am i going wrong?
On
$$\tan(\arctan1+\arctan2+\arctan3)=\frac{1+\tan(\arctan2+\arctan3)}{1-\tan(\arctan2+\arctan3)}=\frac{1+\dfrac{2+3}{1-2\cdot3}}{1-\dfrac{2+3}{1-2\cdot3}}=\frac{1-1}{1+1}=0.$$
This implies
$$\arctan1+\arctan2+\arctan3=k\pi$$ for some $k$.
As the arc tangents are in range $(0,\pi/2)$, we have
$$0<k\pi<\frac{3\pi}{2}$$ which is conclusive.
Note that we have that $$\arctan x +\arctan y =\arctan \frac{x+y}{1-xy} \pmod {\pi}$$ The $\mod \pi$ appears because the tangent function has a period of $\pi$. So it is not necessarily guranteed that $$\arctan x +\arctan y =\arctan \frac{x+y}{1-xy}$$ This is especially true since $\arctan 2$ and $\arctan 3$ are both positive, but $-\frac{\pi}{4}$ is negative in your case. So we should actually have $\arctan 2 +\arctan 3=\frac{3 \pi}{4}$.
Note that $\arctan x=\tan^{-1} x$. I add this so as to avoid confusion in notation.