$\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}$?

This is just another way of saying what the others told you.

$$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3} \ne \lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}$$

The theorem is IF $\displaystyle \lim_{x\to 0}f(x) = L$ and $\displaystyle \lim_{x\to 0}g(x)=M$, where $M, N \in \mathbb R$, THEN $\displaystyle \lim_{x\to 0}(f(x)-g(x))=L-M$

But, since $\displaystyle \lim_{x\to 0} \frac{\tan x}{x^3} = \lim_{x\to 0} \frac{\sin x}{x^3} = \infty$, then the theorem does not apply.

This limit can be evaluated without resorting to L'Hospital.

\begin{align} \frac{\tan x - \sin x}{x^3} &= \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \\ &= \frac{\sin x - \sin x \cos x}{x^3 \cos x} \\ &= \frac{1}{\cos x} \cdot\frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2} \\ &= \frac{1}{\cos x} \cdot\frac{\sin x}{x} \cdot \frac{2\sin^2(\frac 12x)}{x^2} \\ &= \frac{1}{\cos x} \cdot\frac{\sin x}{x} \cdot \frac 12 \cdot \left(\frac{\sin \frac x2}{\frac x2}\right)^2 \\ \end{align}

which approaches $\dfrac 12$ as $x$ approaches $0$.

Your problem arises from the fact that you used $\color{red}{\lim_\limits{x \to 0} \frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $\color{red}{\infty-\infty}$...

Only split an initial limit into a product if the individual limits are defined.

I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ \lim\limits_{x \to 0}\big( f(x) - g(x)\big)$ is not always equal to $ \lim\limits_{x \to 0} f(x) - \lim\limits_{x \to 0} g(x)$.