1. For continuity at $x=0$, we must have $f(0)=\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}.$ So we need to evaluate this limit.

2. Let $u=\tan x, \qquad v=\sin x.$ As $x\to 0$, we have $u\to 0$ and $v\to 0$. Then $\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}=\frac{\tan u-\sin v}{u-v}.$ This is not directly a standard form, so expand near $0$.

3. Use series expansions: $\tan t=t+\frac{t^3}{3}+O(t^5), \qquad \sin t=t-\frac{t^3}{6}+O(t^5).$ Hence $\tan(\tan x)=\tan x+\frac{(\tan x)^3}{3}+O(x^5),$ $\sin(\sin x)=\sin x-\frac{(\sin x)^3}{6}+O(x^5).$ Therefore the numerator is $\tan(\tan x)-\sin(\sin x)=(\tan x-\sin x)+\frac{(\tan x)^3}{3}+\frac{(\sin x)^3}{6}+O(x^5).$ So $f(x)=1+\frac{\frac{(\tan x)^3}{3}+\frac{(\sin x)^3}{6}+O(x^5)}{\tan x-\sin x}.$

4. Now expand $\tan x-\sin x$: $\tan x=x+\frac{x^3}{3}+O(x^5),$ $\sin x=x-\frac{x^3}{6}+O(x^5).$ Thus $\tan x-\sin x=\frac{x^3}{3}+\frac{x^3}{6}+O(x^5)=\frac{x^3}{2}+O(x^5).$ Also, $(\tan x)^3=x^3+O(x^5), \qquad (\sin x)^3=x^3+O(x^5).$ Hence $\frac{(\tan x)^3}{3}+\frac{(\sin x)^3}{6}=\frac{x^3}{3}+\frac{x^3}{6}+O(x^5)=\frac{x^3}{2}+O(x^5).$

5. Substitute into the expression for $f(x)$: $f(x)=1+\frac{\frac{x^3}{2}+O(x^5)}{\frac{x^3}{2}+O(x^5)}.$ Therefore, $\lim_{x\to 0} f(x)=1+1=2.$

6. Since $f(x)$ is continuous at $x=0$, $f(0)=2.$

Thus the required integer is $\boxed{2}$.

Comparison with stored answer: the stored correct answer is $2$, which matches our derived result.