We start with the expression

${\displaystyle \tan (A-B)+\tan (B-C)+\tan (C-A)=0.}$

Write $X=A-C$ and $Y=B-C$, then the given expression becomes

${\displaystyle \tan (X-Y)+\tan Y+\tan -X=0.}$

This gives

${\displaystyle \tan (X-Y)=\tan X-\tan Y}$

and we know the identity

${\displaystyle \tan (X-Y)=\frac{\tan X-\tan Y}{1-\tan X\tan Y}.}$

Hence either

${\displaystyle \tan X=\tan Y\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1)}$

or

${\displaystyle \tan X\tan Y=0\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(2)}$

In case (1) we show that the angles $X$ and $Y$ are equal.

${\displaystyle |X-Y|=|A-B|<A+B<{180}^{\circ }}$

and the tan function is periodic with period 180 degrees so $X=Y.$ This gives $A-C=B-C$ hence $A=B$, so the triangle is isosceles.

In case (2), either $\tan X=0$ or $\tan Y=0$, hence $A=C$ or $B=C$ and in all the cases the triangle is isosceles.