Problem on continuity based on functional relation Given that  $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfies 
$2f^3(x)-3=2x-3f(x)$ , $x\in \mathbb{R}$, show that  $f$ is continuous on $\mathbb{R}$.
How can we handle this problem?
 A: Let us define $$F(x):=2f^3(x)+3f(x)$$
then
$$F(x)=3+2x$$
hence $F$ is continuous. Now let us define $$ G(y):= 2y^3 + 3y $$
then, since $G'(y)=6y^2+3>0\ $, (plus computing the limits in $\pm\infty$, $G:\mathbb{R}\rightarrow \mathbb{R}$ is a diffeomorphism (i.e. $t\mapsto G^{-1}$ exists and is continuous).
We have 
$$ F=G\circ f$$
so 
$$ f=G^{-1}\circ F$$
hence $f$ is continuous.
A: The solution  of @Netchaiev is short and smart. I managed to get to  a lengthy alternative below:
For $x_0 \in \mathbb{R}$ we have $2f^3(x_0)=2x_0-3f(x_0)+3$
Subtract  the latter  from  the given $2f^3(x)=2x-3f(x)+3$ to take
$2(f^3(x)-f^3(x_0)) +3(f(x)-f(x_0))= 2(x-x_0)$
or
$2(f(x)-f(x_0))(f^2(x)+f(x)f(x_0)+f^2(x_0))+3(f(x)-f(x_0))=2(x-x_0)$
or
$(f(x)-f(x_0))(2f^2(x)+2f(x)f(x_0)+2f^2(x_0)+2+1)=2(x-x_0)$
Let $A(x) = 2f^2(x)+2f(x)f(x_0)+2f^2(x_0)+2$
the above is quadratic with respect to $f(x)$ with discriminant 
$\Delta = (2f(x))^2-4\cdot2\cdot(2f^2(x)+2)=4f^2(x)-8\cdot2\cdot f^2(x)-16= -12f^2(x)-16<0$
thus $A(x)>0, \forall x$ and 
$(f(x)-f(x_0))(A(x)+1)=2(x-x_0)$. Then 
$\left|f(x)-f(x_0)\right|=\frac{\left|2(x-x_0)\right|}{A(x)+1}\leq \left|2(x-x_0)\right|$, since $A(x)+1>1$
or
$-\left|2(x-x_0)\right| \leq f(x)-f(x_0) \leq \left|2(x-x_0)\right|$
We have that $\lim_{x \to x_0} 2(x-x_0) = 0$
then from the squeeze theorem it follows that $\lim_{x \to x_0} (f(x)-f(x_0)) = 0$, thus $f(x)$ is continuous
