Paso 1: Usar la identidad notable:
$$ \frac{a^{7}-b^{7}}{a-b}=a^{6}+a^{5}b+a^{4}b^{2}+a^{3}b^{3}+a^{2}b^{4}+ab^{5}+b^{6}. $$
Tomando $a=\sen x$ y $b=\cos x$:
$$ E=\sum_{k=0}^{6}(\sen x)^{6-k}(\cos x)^k. $$
Esta expresión es simétrica en $\sen x$ y $\cos x$ y depende de:
$$ p=\sen x+\cos x,\qquad q=\sen x\cos x. $$

Paso 2: De $\tan x+\cot x=3$:
$$ \tan x+\cot x=\frac{\sen x}{\cos x}+\frac{\cos x}{\sen x} =\frac{\sen^{2}x+\cos^{2}x}{\sen x\cos x} =\frac{1}{\sen x\cos x}=3 \Rightarrow q=\sen x\cos x=\frac{1}{3}. $$

Paso 3: Hallar $p^{2}$:
$$ p^{2}=(\sen x+\cos x)^{2}=\sen^{2}x+\cos^{2}x+2\sen x\cos x =1+2\cdot\frac{1}{3}=\frac{5}{3}. $$

Paso 4: Usar la recurrencia (polinomios simétricos homogéneos):
Si $h_0=1,\;h_1=p$ y
$$ h_n=p\,h_{n-1}-q\,h_{n-2}, $$
entonces $E=h_6$.

Calculando (y simplificando) se obtiene:
$$ h_6=p^{6}-\frac{5}{3}p^{4}+\frac{2}{3}p^{2}-\frac{1}{27}. $$
Sustituyendo $p^{2}=\frac{5}{3}$:
$$ p^{6}=\left(\frac{5}{3}\right)^{3}=\frac{125}{27},\quad p^{4}=\left(\frac{5}{3}\right)^{2}=\frac{25}{9}. $$
Entonces:
$$ E=\frac{125}{27}-\frac{5}{3}\cdot\frac{25}{9}+\frac{2}{3}\cdot\frac{5}{3}-\frac{1}{27} =\frac{125}{27}-\frac{125}{27}+\frac{10}{9}-\frac{1}{27}. $$
$$ E=\frac{10}{9}-\frac{1}{27}=\frac{30}{27}-\frac{1}{27}=\frac{29}{27}. $$

Resultado final: $\boxed{\frac{29}{27}}$.