Reciprocal squares of x and tangent

Step 1: the expansion of $\tan x$. As $x\to0$,
$$\begin{aligned} \tan x&=\frac{\sin x}{\cos x}=\frac{x-\frac{x^3}6+o(x^3)}{1-\frac{x^2}2+o(x^2)}\\&=\left(x-\frac{x^3}6\right)\left(1+\frac{x^2}2\right)+o(x^3)\\&=x+\frac{x^3}3+o(x^3). \end{aligned}$$

Step 2: combine.
$$\frac1{x^2}-\frac1{\tan^2x}=\frac{\tan^2x-x^2}{x^2\tan^2x}=\frac{(\tan x-x)(\tan x+x)}{x^2\tan^2x}.$$

Step 3: orders of magnitude. From Step 1:
- $\tan x-x=\dfrac{x^3}3+o(x^3)$,
- $\tan x+x=2x+o(x)$,
- $x^2\tan^2x=x^4+o(x^4)$.

So the numerator is $\dfrac{2x^4}3+o(x^4)$ and the quotient is
$$\frac{\frac23x^4+o(x^4)}{x^4+o(x^4)}\longrightarrow\frac23.$$

Remark. The naive attempt "$\tan x\approx x$, so the difference is $0$" gives the wrong answer: the information is hidden in the $x^3$ term of $\tan x$.