Does there exist a function $f$ such that $\forall x \in \mathbb{R}:f^3(x) + f^2(x) \cdot x^2 = 1$?
I haven't studied functional equations so I have no idea how to solve this problem. I think I proved it's impossible if $f$ is polynomial (it would have to be $f(x) = 1 - x^2$, but that doesn't work). But what I really want to do is the opposite, I want to find $f$ with this property, as it would work as a counter-example to some phenomena I'm investigating about derivatives.
Is it possible?
Assuming that we face the problem of $$y^3+x^2y^2-1=0 \qquad \text{with} \qquad y=f(x)$$ following the steps given here, we have $$\Delta=4 x^6-27 \qquad p=-\frac{x^4}{3}\qquad q=\frac{2 x^6}{27}-1$$
If $\Delta <0$, using the hyperbolic method for only one real root $$y=\frac{1}{3} x^2 \left(2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{27}{2 x^6}-1\right)\right)-1\right)$$
If $\Delta > 0 $, the equation has then three real solutions given by
$$y_k=\frac{1}{3} x^2 \left(2 \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(\frac{27}{2 x^6}-1\right)\right)\right)-1\right)$$ with $k=0,1,2$.