Prove that there is a unique function of class $C^1$ such that for all $x \in \mathbb{R}$: $x^2f(x)^3 + 3x^3f (x)^2 + (5x^4 + 1) f (x) = \cos (f (x))$ Prove that there is a unique function $f: \mathbb{R} \to \mathbb{R}$ of class $C^1$ such that for all $x \in \mathbb{R}$: 
$$x^2f(x)^3  + 3x^3f (x)^2 + (5x^4 + 1) f (x) = \cos (f (x))$$
My work: 
Let $y=f(x)$ and by using implicit theorem is idea to solve this:
$$F(x,y)=x^2y^3  + 3x^3y^2 + (5x^4 + 1)y - \cos (y)=0=F(x,f(x))$$
(F is class $C^1$)
To show this we must show that: $\frac{\partial F}{\partial y}$ is regular matrix for all x.
$$\frac{\partial F}{\partial y}(x,y)=3y^2x^2+6yx^3+5x^4+1-\sin(y)$$
$$\frac{\partial F}{\partial y}(x,y)=3y^2x^2+6yx^3+5x^4+1-\sin(y)=0$$ 
But for $(x,y)=(0,\pi/2)$ $F$ is 0 - which isn't good :/
What I did wrong? 
 A: 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)
