You can take the derivative to show that it is always negative except at $0$ where it's zero. Though that's not obvious, so you take the derivative again to show the same thing. Until you get something obvious, then work your way back up.
Ok, I think you got it. So I'll add the answer now.
The first three derivatives are:
$$f'(x)=3 x^2 - 2\sin^2(x) - \tan^2(x)$$ $$f''(x)=6 x - 4\cos(x)\sin(x) - 2\sec^2(x)\tan(x)$$ $$f'''(x)=-2(5+2\cos(2x))\tan^4(x)$$
Just plug and play to see that $f(0)=0$, $f'(0)=0$, and $f''(0)=0$.
You can see that $5+2\cos(2 x)\ge 3$ and $\tan^4(x)\ge 0$. Therefore $f'''(x)\le 0$.
As long as $f'''(x)$ is continuous, which is true up to $x<{\pi\over 2}$, then we have $f''(x)<0$ for $0<x<{\pi\over 2}$.
Rinse and repeat. $f'(x)$ is zero at zero and then decreasing, since $f''(x)$ is negative after zero. So $f'(x)<0$ for $0<x<{\pi\over 2}$. Now $f(x)$ is zero at zero and decreasing, since $f'(x)$ is negative after zero.
Therefore $f(x)<0$ for $0<x<{\pi\over 2}$.
Update:
@Aryabhata asked how I got that expression for the third derivative. It was by factoring a polynomial. Without any messing around, you get for the third derivative:
$$6 - 4\cos^2(x) + 4\sin^2(x) - 2\sec^4(x) - 4\sec^2(x)\tan^2(x)$$
You then convert everything to sines, where here we replace $\sin(x)$ with $s$:
$$6 - 4(1-s^2) + 4 s^2 - {2\over (1-s^2)^2} - {4\over (1-s^2)}{{s^2}\over(1-s^2)}$$
Now you pull out a common numeric factor and denominator, and then expand to get:
$$-{2\over(1-s^2)^2}(7 s^4 - 4 s^6)$$
and factoring:
$$-{2 s^4\over(1-s^2)^2}(7 - 4 s^2)$$
There are some obvious trig identities we can apply to simplify that, but for the purposes of this problem, we'll stop right there. We can already see by inspection that $f'''(x)\le 0$, which is all that is needed for the answer above.
(Note: when I did this problem originally, I replaced everything with cosines instead of sines. That made factoring the polynomial much harder. When I saw the answer, I realized I should have used sines instead. So when I wrote up this answer, I used sines to make myself look smarter than I really am. We all do that when we write up proofs, right? We look for the shortest path, probably taking $17$ different dumb detours before writing the final elegant-looking solution. Or maybe that's just me.)