According to Rudin,
The inverse function theorem states, roughly speaking, that a continuously differentiable mapping $f$ is invertible in a neighborhood of any point $x$ at which the linear transformation $f'(x)$ is invertible.
Given a function $f$, how do you find the inverse around the point? In a problem similar but not identical to my homework,
let $f=(f_1,f_2)$ be the mapping from $R^2$ into $R^2$ given by $f_1(x,y)=e^xcos(y)$, $f_2(x,y)=e^xsiny$. Put $a=(0,\frac{π}{3})$, $b=f(a)$. Let $g$ be the continuous inverse of f, defined in a neighborhood of b, such that $g(b)=a$. Find an explicit formula for g and calculate $f'(a)$ and $g'(b)$.
In general, it cannot be done. For instance, if $f$ is the mapping from $\Bbb R$ to itself given by $$ f(x) = x^5 + x + 1 $$ then near $x = 2, f(x) = 35$, there's a well-defined inverse...but you're not going to be able to write it down explicitly in terms of things like square roots, etc.
But more generally, for a function from 2-space to itself, say $$ f(x, y) = (2x + 3y, 3x - y) $$ you can write $$ (u, v) = (2x + 3y, 3x - y) $$ so that $$ u = 2x + 3y\\ v = 3x - y $$ and then you try to solve for $x$ and $y$ in terms of $u$ and $v$. For this example, you might multiply the equations by $3$ and $2$ respectively to get $$ 3u = 6x + 9y\\ 2v = 6x - 2y $$ and then subtract to get $$ 3u - 2v = 11 y $$ whence $$ y = \frac{3u - 2v}{11}, $$ and you can do something similar for $x$.
The tools you use (like my choice to multiply by 3 and 2) will depend on the equations given, and you have to guess what'll allow you to eliminate one variable or the other.