Proving that $\frac{6\tan\phi}{\tan^2\phi-9}=\tan A$

In a triange $ABC$, points $D$ and $E$ are taken on side $BC$ such that $BD=DE=EC$. If $\angle ADE=\angle AED=\phi$, how can we prove that
$$\frac{6\tan\phi}{\tan^2\phi-9}=\tan A$$

The key here is recognizing the identity we wish to prove. Note that

$$\tan{A} = -\frac{\frac{2}{3} \tan{\phi}}{1-(\frac13 \tan{\phi})^2}$$

This looks like a double-angle formula for tangent. Now note that, because of symmetry, $B=C$. Thus $A = \pi-2 B$ and $\tan{A} = -\tan{2 B}$. Therefore, the formula implies that

$$\tan{B} = \frac13 \tan{\phi}$$

This is what we should need to prove. We do this by noting two items: 1) The triangle $\Delta ABE$ is isosceles with side $AD=AE=d$ given by

$$d \cos{\phi} = \frac{a}{6}$$

2) The law of sines in either side triangle, say $\Delta ABD$ produces the equation

$$\frac{\sin{(\phi-B)}}{\sin{B}} = \frac{a}{3 d}$$

or

$$\frac{\sin{\phi}}{\tan{B}} - \cos{\phi} = 2 \cos{\phi}$$

The above assertion - and consequently, the original relation - is proven.