Step 1: Turn big angles into small ones.
Use $\tan(90^\circ-\theta)=\cot\theta.$ So $\tan81^\circ=\cot9^\circ$ and $\tan63^\circ=\cot27^\circ.$
Step 2: Rewrite the expression.
The expression becomes \[ E=\cot9^\circ-\cot27^\circ-\tan27^\circ+\tan9^\circ. \]
Step 3: Group matching pairs.
Group as $(\cot9^\circ+\tan9^\circ)-(\cot27^\circ+\tan27^\circ).$ Wait, signs: we have $\cot9^\circ+\tan9^\circ$ and $-(\cot27^\circ+\tan27^\circ).$
Step 4: Use a helpful identity.
Note $\cot\theta+\tan\theta=\dfrac{\cos\theta}{\sin\theta}+\dfrac{\sin\theta}{\cos\theta}=\dfrac{1}{\sin\theta\cos\theta}=\dfrac{2}{\sin2\theta}.$ So each pair becomes $\dfrac{2}{\sin2\theta}.$
Step 5: Apply it.
$E=\dfrac{2}{\sin18^\circ}-\dfrac{2}{\sin54^\circ}.$ Using exact values $\sin18^\circ=\dfrac{\sqrt5-1}{4}$ and $\sin54^\circ=\dfrac{\sqrt5+1}{4}$, this becomes $\dfrac{8}{\sqrt5-1}-\dfrac{8}{\sqrt5+1}.$
Step 6: Simplify the difference.
Common denominator $(\sqrt5)^2-1=4$: $E=\dfrac{8(\sqrt5+1)-8(\sqrt5-1)}{4}=\dfrac{16}{4}=4.$ \[ \boxed{4} \]