Can you solve any mathematical function? For any finite mathemathical function (consisting of addition, subtraction, division, multiplication, exponentiation, trigonometry) can you find $x$ in $f(x) = y$ where $y$ is a number you want?
Is it proven that any function is solvable or not solvable? Can a computer solve a function that's millions of characters long?
I tried to solve a large function in maxima but it got in an infinite loop.
 A: An elementary function  need not have an elementary inverse. Consider
$$f(x)=xe^x,$$
whose inverse is the Lambert W function, which cannot be expressed in terms of elementary functions (see here).
A: As Golden_Ratio noted, there isn't always a nice, closed form for the inverse of an elementary function. But a computer can get arbitrarily close to a solution to an elementary function, if it exists. Have a look at the Wikipedia page on root-finding algorithms.
A: I suppose you mean solving a function ($y$ is a variable), not solving a number equation ($y$ is a number). That are two different mathematical problems.
Your problem makes sense only if $x$ is a variable for elements of the domain of the function $f$ and $y$ is a variable for elements of the range of the function $f$. If $x$ is not a variable for elements of the domain of $f$ or $y$ is not a variable for elements of the range of $f$, the function is not solvable.
A function is a relation between two sets (domain and range) that assigns exactly one element of the range to each element of the domain. An inverse function exists only for bijective functions, but the inverse relation always exists. Your question asks for determining the inverse relation of $f$. If you are looking for functions instead of a relation, you can subdivide the inverse relation into its branches (function branches), the partial inverses (means partial inverse functions) of $f$.
Because the inverse relation always exists, it can be calculated numerically.
You can also ask for partial inverses in closed form.
If $f$ is a composition of functions $f_1,...,f_n$: $f(x)=f_n(f_{n-1}(...(f_2(f_1(x)))))$ and the partial inverses of $f_1,...,f_n$ are known/allowed, the partial inverses $f^{-1}$ of $f$ are: $f^{-1}(x)=f_1^{-1}(f_2^{-1}(...(f_{n-1}^{-1}(f_n^{-1}(x)))))$, where $f_i^{-1}$ is a partial inverse of $f_i$ ($i\in\{1,...,n\}$).
Let $A$ be an algebraic function and $f_1,...,f_n$ algebraically independent functions. You cannot solve $A(f_1(x),...,f_n(x))=y$ for $x$ if the domain of the function $A\circ f_1\circ ...\circ f_n$ is a non-discrete set and simultaneously only $A,f_1,...,f_n$ and their partial inverses are known/allowed.
$f$ is an explicit elementary function in your example.
For elementary partial inverses of elementary functions, the algebraic independence is proved implicitly in the main theorem in [Ritt 1925] that is also proved in [Risch 1979]:
If $f$ and its inverse are both elementary, then $f(z)=\phi_n\circ\  ...\circ\ \phi_1(z)$, where each $\phi_i$ is either an algebraic function of $z$ or else $\exp$ or $\ln$.
See e.g. the function $f$ with $f(x)=xe^x$ from the other answer.
$\ $
[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90
[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759
