$f(x,y)=(\sin x e^y, \cos x e^y)$- this is continuous. How would I prove that it's inverse is continuous as well? 
$f(x,y)=(\sin (x) e^y, \cos (x) e^y)$- this is continuous. How would I prove that it's inverse is continuous as well?

I need this for inverse function differentiating theorem that says that f has to be a homomorphism akka a bijection and its inverse has to be a bijection. Now if $f$ is is bijective doesnt that make it an isomorphism as well? That definition is wierd to me..
 A: Let $u = e^y\sin x, v = e^y\cos x\Rightarrow u^2+v^2 =e^{2y}\Rightarrow y = \dfrac{\ln(u^2+v^2)}{2}, \dfrac{u}{v} = \tan x\Rightarrow x = \tan^{-1}\left(\dfrac{u}{v}\right)\Rightarrow f^{-1}(x,y) =\left(\tan^{-1}\left(\dfrac{x}{y}\right),\dfrac{\ln(x^2+y^2)}{2}\right)$. From this you can prove that $f^{-1}$ is continuous.
A: For a more convenient notation, take $f(x,y)=e^x(\cos y,\sin y)$. If you think about it for a second, this is the complex exponential function that sends $z=x+iy$ to $$e^z=e^{x+iy}=e^xe^{iy}=e^x(\cos y+i\sin y)$$ Then the Jacobian determinant of $f$ is $|f'(z)|^2=e^{2x}$ which is nonzero everywhere. The inverse function theorem then guarantees $f$ is locally a diffeomorphism. Note that $f$ cannot be bijective since it is periodic in the second coordinate, in particular it is not injective. It is also not surjective since it misses the origin. 
A: If we look at $f$ as a map from $\mathbb{C}$ to $\mathbb{C}$, by setting $z=x+iy$ and:
$$g(z) = g(x+iy) = e^y \sin x + i e^y \cos x = i\,e^{y-ix} = i e^{-iz}, $$
it is trivial that we are discussing the local invertibility of the complex exponential (an entire function with a nowhere-vanishing derivative) and the continuity of (a branch of) the complex logarithm away from zero. Obviously there is no global inverse since $g(z+2\pi)=g(z)$.
