Inverse of the composition of two functions If I have a composition of two functions:
$$y = f(g(x),h(x))$$
where both $g(x)$ and $h(x)$ are readily invertible, can I find the inverse of the composition? i.e.: Can I find $x = f^{-1}(y)$? I know this is generally possible for the the composition of one function. 
Perhaps there is a special form of f that permits this?
Thanks
 A: EDIT: Previous answer used the accepted definition of 'composition'. I am trying to interpret what the OP means by composition... I think he means that $f$ is a formula e.g. something like $f(\sin x,e^x)=\sin x\sqrt{e^x}$ means that $f(x,y)=x\sqrt{y}$.
Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$, $g:\mathbb{R}\rightarrow \mathbb{R}$ and $h:\mathbb{R}\rightarrow \mathbb{R}$ be invertible maps.
Define a function $F:\mathbb{R}\rightarrow \mathbb{R}$ by
$$y=F(x)=f(g(x),h(x)).$$
Then $F$ is invertible with
$$x=F^{-1}(y)=g^{-1}\left(\pi_1\left(f^{-1}(y)\right)\right)=h^{-1}\left(\pi_2\left(f^{-1}(y)\right)\right),$$
where $\pi_i:\mathbb{R}^2\rightarrow\mathbb{R}$ is the projection onto the $i$-th factor of $\mathbb{R}^2$.
A: An answer for the general case is given by the previous answer.
We write
$$F(x)=f(g(x),h(x)).$$
a)
Let's consider the case that your function $f$ is an algebraic function. We write $A=f$.
We presuppose $g,h$ are invertible functions over an open domain. We denote their inverses by $g^{-1}$ and $h^{-1}$ respectively.
Let's investigate the equation
$$A(g(x),h(x))=y,\tag{1}$$
where $g(x),h(x)$ are expressions and $A(g(x),h(x))$ is an algebraic expression in dependence of both $g(x)$ and $h(x)$ at the same time. So $A$ is an algebraic function with $A\colon D\subseteq\mathbb{C}^2\to\mathbb{C}$. According to the invariance of dimension, $A$ is not injective and therefore not invertible.
But if the functions $g,h$ are algebraically dependent, an algebraic function $A_0$ exists so that $h(x)=A_0(g(x))$. Equation (1) then becomes
$$A(g(x),A_0(g(x)))=y.$$
Because the composition of algebraic functions is an algebraic function, an algebraic function $A_1$ then exists so that
$$A_1(g(x))=y.$$
If $A_1$ is bijective, we get
$$x=g^{-1}(A_1^{-1}(y)).$$
The main theorem in [Ritt 1925] that is also proved in [Risch 1979] applies this situation for the Elementary functions.
The Elementary functions (according to Liouville and Ritt) are those functions which are obtained in a finite number of steps by performing algebraic operations and taking exponentials and logarithms. If your functions $f$, $g$ and $h$ are elementary functions, a result of the main theorem in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 is: your elementary function $F$ can have an inverse that is an elementary function only if there exists an exp/ln-composition representation of $F$ whose member functions are all 1-ary functions.
Because each function is composed of bijective pieces, the problem can be applied to partial inverses.
b)
If an existing inverse function is accepted, depends on the set of functions you allow. Take e.g.
$$F(x)=f(g(x),h(x))=x\cdot e^x.$$
We have:
$F\colon x\mapsto x\cdot e^x$,
$f\colon (x_1,x_2)\mapsto x_1\cdot x_2$,
$g\colon x\mapsto x$,
$h\colon x\mapsto e^x$ and
$F^{-1}\colon x\mapsto \textrm{LambertW}(x)$.
LambertW is a known funtion, but it's not an elementary function.
$\ $
[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759
[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90
