Derive $\tan(3x)$ in terms of $\tan(x)$ using De Moivre's theorem 
Derive the following identity: $$\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$$

The way I approached the questions is that I first derived $\sin(3x)$ and $\cos(3x)$ because $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$. Then substituting:
$$\tan(3x)=\frac{\sin(3x)}{\cos(3x)}=\frac{-4\sin^3(x)+3\sin(x)}{4\cos^3(x)-3\cos(x)}$$
I transformed it into $$\tan(3x)=\frac{-4\sin^2(x)\tan(x)+3\tan(x)}{4\cos^2(x)-3}$$ and also to $$\tan(3x)=-\tan(x)\frac{4\sin^2(x)-3}{-4\sin^2(x)+1}$$ but I'm stuck on either of these forms. Any help?

Don't “normalize” the triplication formulas to only sines and cosines.
We have
$$
\cos3x+i\sin3x=\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x
$$
so
\begin{align}
\sin3x&=3\cos^2x\sin x-\sin^3x \\[6px]
\cos3x&=\cos^3x-3\cos x\sin^2x \\[12px]
\tan3x&=\frac{3\cos^2x\sin x-\sin^3x}{\cos^3x-3\cos x\sin^2x}
\end{align}
Now divide numerator and denominator by $\cos^3x$.
