Prove that $\tan(\sin(x))=x-\frac{x^{3}}{6}+o(x^{3})$ 
Prove that $\tan(\sin(x))=x-\frac{x^{3}}{6}+o(x^{3})$

I know that when I calculate derivative then I get true answer. However I want to know why the way that I will present soon does not work. My try:$$\tan(x)=x+r_{1}(x), \quad  r_{1}(x)=o(x)$$ $$\tan(\sin(x))=\sin(x)+r_{1}(\sin x),\quad r_{1}(\sin x)=o(\sin x)$$ $$\tan(\sin(x))=x-\frac{x^{3}}{6}+r_{2}(x)+r_{1}(x-\frac{x^{3}}{6}+r_{2}(x)),\quad r_{2}(x)=o(x^{3})$$ In this moment I should find: $$o(?)=r_{1}(x-\frac{x^{3}}{6}+r_{2}(x))$$I suspect that $o(?)=o(x^{3})$ so I write: $$\frac{r_{1}(x-\frac{x^{3}}{6}+r_{2}(x))}{(x-\frac{x^{3}}{6}+r_{2}(x))}\cdot \frac{(x-\frac{x^{3}}{6}+r_{2}(x))}{x^{3}}$$ But in this moment I have $0 \cdot (+\infty)$. Moreover I see that $$r_{1}(x-\frac{x^{3}}{6}+r_{2}(x))=o(x)$$ because $$\frac{r_{1}(x-\frac{x^{3}}{6}+r_{2}(x))}{(x-\frac{x^{3}}{6}+r_{2}(x))}\cdot \frac{(x-\frac{x^{3}}{6}+r_{2}(x))}{x} \rightarrow 0$$Can someone tell me where I'm making a mistake, that I'm getting $ o (x) $ instead of $ o (x ^ {3}) $?

The problem statement contains a sign error.
In any case you want to compute $\tan \sin x$ modulo $o(x^3)$, so you should compute $\sin x$ and $\tan x$ modulo $o(x^3)$, not $o(x)$. Indeed,
$$\sin x = x - \frac{1}{6} x^3 + o(x^3), \qquad \tan y = y + \frac{1}{3} y^3 + o(y^3) ,$$
so substituting gives $$\color{#df0000}{\boxed{\tan \sin x = \left(x - \frac{1}{6} x^3\right) + \frac{1}{3} \left(x - \frac{1}{6} x^3\right)^3 + o(x^3) = x + \frac{1}{6} x^3 + o(x^3)}} .$$
