In order to apply the implicit function theorem you first need to find a point $(x,y)$ such that $F(x,y) =0$. One possible approach is to use the mean value theorem. You may for example check that $F(x, 0) = - \cos(0) = -1$ is independent of $x$. Now $$F(x,\frac{\pi}{2}) = x^2 \left(\frac{\pi}{2}\right)^3 + 3 x^3 \left(\frac{\pi}{2}\right)^2 + (5x^4 +1)\frac{\pi}{2} $$
is, quite obviosly, positive for large $x$. If you fix such an $x$, $x_0$, say, then the mean value theorem for continuous functions shows there is (at least) one value $y_0$ such that $F(x_0, y_0) = 0$.
You now have to check that the $y$- derivative of $F$ does not vanish when $F(x, y) = 0$ There is no need to check this where this condition does not hold.
This allows you to locally solve $F(x,y)$ uniquely in a neighbourhood of $(x_0,y_0)$.
You then need to show that you can extend the solution (uniquely) to $x\in \mathbb{R}$. This is usually done by assuming the solution exists for $x \in (a,b)$ an the let $x\rightarrow a$ (or $b$) and check whether the resulting $F(x, f(x))$ converges. If it does, it will converge to $0$. Then, if $\frac{\partial}{\partial y}F(x,f(x))$ is not zero at that point, you can (uniquely) extend the solution. So the set where $f$ is defined is both open (why?) and closed, so it's connected so it's $\mathbb{R}$.
If you arrived here, you are, however, not yet done. You have to exclude the possibility that there is a second solution. One starting point is to check whether $F(x, y) $ may be strictly monotonic somewhere with respect to $y$. If it is, and if your first considerations show that a local solution extends to a global one in a unique way, then the assumption that a second solution exists would be a contradiction.
The details I will leave to you (or some other kind soul who likes to write down the details or knows a shortcut).
(oh, and your last calculation shows that $F(0, \pi/2) \neq 0$, which is meaningless here, as indicated already by coffeemath)
